# NCERT Solutions Class 7 Maths Chapter 7

## NCERT Solutions for Class 7 Mathematics Chapter 7 Congruence of Triangles

Triangles that have the same size and shape are called congruent triangles. The symbol ≅ is used to indicate the congruence between them. Since this is an important and interesting  topic, Extramarks offers NCERT Solutions for Class 7 Mathematics Chapter 7 that will encourage  and guide you through the topic. These are detailed step-by-step solutions that cover all the questions covered in this chapter. Students will find these solutions very helpful for their exam preparation.

### NCERT Solutions for Class 7 Mathematics Chapter 7 – Congruence of Triangles

 Chapter 7 – Congruence of Triangles Exercises Exercise 7.1 Questions & Solutions Exercise 7.2 Questions & Solutions

## NCERT Solutions for Class 7 Mathematics-

Congruence is the term usually used in Mathematics to define an object and its mirror image. Two objects or shapes are said to be congruent to each other if they superimpose on each other. In other words, we can say that their shape and dimensions are the same. In the case of geometric figures, line segments are congruent if they are of the same length and angles are congruent if they are of the same measure.

In the case of triangles, the corresponding sides, as well as the angles of congruent triangles, are all equal. There are certain criterias to check  if two triangles are congruent or not. It is advisable to first find out the dimensions of the triangles before trying to prove them congruent. However, the evaluation of the congruence of triangles can be done by proving only three values out of these six.

The chapter covers the following topics:

•  Congruence of Angles
• Congruence of Plane Figures
• Congruence Among Line Segments
• Congruence of Triangles
• Criteria for Congruence  of Triangles

Some Facts of Angles

• The vertically opposite angles are always equal.
• Two adjacent angles are said to form a linear pair only if their sum is 1800.
• The alternate angles and corresponding angles are equal when two parallel lines are intersected by a transversal.
• The sum of interior opposite angles on the same side of the transversal is always 1800.
• In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of its other two sides.

Congruence of Plane Figures

When a plane figure covers the other one completely, then they are said to be of the same shape and same size (such objects are said to be congruent). Thus, we can say that congruent objects are replicas of one another and the relationship between such congruent objects is termed Congruence.

Note: As per the rule of congruence, two congruent objects are always similar but two similar figures are not always congruent, they may or may not be congruent.

For Example:

1. Any two squares can be similar but they are congruent only if they are of the same length.
2. Any two circles are always similar but they are congruent only if they are of the same radius.
3. Any two equilateral triangles can be similar but they are congruent only if the length of their sides is the same.

Note:

If two line segments are of the same length, they are said to be congruent. Also, if two line segments are congruent, it implies that they have the same length.

If two angles are of the same measure, they are said to be congruent. Also, when two angles are congruent, it implies their measure is the same.

Thus, we can say that the congruence of angles depends on their measures i.e. if they are equal or not.

To indicate that two line segments are congruent, the symbol ≅ is used.

The same symbol is used for congruent angles. For example, if ∠ABC and ∠PQR are congruent, it will be written as ∠ABC ≅∠PQR.

Congruence of Triangles

Two triangles are said to be congruent if they are copies of each other and cover each other exactly if superimposed.

For example, consider two triangles: ΔABC and ΔPQR. If both of them are congruent to each other, we can write them as ΔABC ≅ ΔPQR.

Since the literal meaning of congruent is “equal in all respects”, a triangle is congruent to another triangle when both of them are identical or equal to each other in all respects. In other words, when one of them is placed on the other, they should coincide with each other exactly. Only then they can be called congruent to one another.

Thus, any two triangles are said to be congruent if all of their six elements, including three sides and three angles of triangles are equal to the corresponding six elements of the other.

Criteria For Congruence Of Triangles

The congruence of triangles can be proved even without the actual measure of the sides and angles of the triangles. The different rules of congruency include:

1. SSS (Side-Side-Side)
2. SAS (Side-Angle-Side)
3. ASA (Angle-Side-Angle)
4. AAS (Angle-Angle-Side)
5. RHS (Right angle-Hypotenuse-Side)

They have been discussed in detail below:

1. SSS (Side-Side-Side)- If all the three sides of a triangle are equal to the corresponding three sides of the other triangle, then the two of them are said to be congruent by the SSS rule.
2. SAS (Side-Angle-Side)- If any two sides of a triangle and an angle included between the sides of such a triangle are equal to the corresponding two sides and the angle between the sides of another triangle, then the two of them are said to be congruent by the SAS rule.
3. ASA (Angle-Side- Angle)- If any two angles and a side included between the angles of a triangle are equal to the corresponding two angles and side included between the angles of another triangle, then the two of them are said to be congruent by the ASA rule.
4. AAS (Angle-Angle-Side)- When two angles of a triangle and a non-included side of a triangle are equivalent to the corresponding angles and sides of another, then the triangles are said to be congruent by the AAS rule.

Here’s an example of how to prove congruence by the AAS rule. Suppose there are two triangles ABC and XYZ, where

∠B = ∠Y [Corresponding sides] ∠C = ∠Z [Corresponding sides] and

By angle sum property of triangle, we know that;

∠A + ∠B + ∠C = 180 degree

∠X + ∠Y + ∠Z = 180 degree

From equations 1 and 2 we can say;

∠A + ∠B + ∠C = ∠X + ∠Y + ∠Z

∠A + ∠E + ∠F = ∠X + ∠Y + ∠Z [Since ∠B = ∠Y and ∠C = ∠Z] ∠A = ∠X

Hence, in triangles ABC and XYZ,

∠A = ∠X

AC = XZ

∠C = ∠Z

Hence, by the ASA rule of congruence,

Δ ABC ≅ Δ XYZ

1. RHS (Right angle-Hypotenuse-Side)- If the hypotenuse and a side of a right-angled triangle are equal to the hypotenuse and a side of the other right-angled triangle, then the two of them are said to be congruent by the RHS rule.

### NCERT Solutions for Class 7 Mathematics

In case you found the solutions of NCERT Class 7 Mathematics Chapter 7 useful and want to continue studying Mathematics through Extramarks website, you can click on the link below to find solutions for the other chapters.

## NCERT Solutions for Class 7

Studying through study material along with the solutions helps students in developing a better understanding of their core subjects. It provides you with an in-depth comprehension of the course and encourages you to practice more by making the concepts easier. It completely eliminates any possibility of  subjects being challenging, perhaps even giving them a  unique perspective.

The NCERT Solutions of Extramarks for Class 7 cover all the topics of Class 7 in a detailed and comprehensive manner. With the assistance of these up-to-date solutions, studying for Class 7 is made easier.  So in case you are interested in solutions for all the subjects of Class 7, you can click on the link given below.

NCERT Solutions Class 7 Maths Chapter-wise List

Q.1

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B, AB}=\text{24 cm, BC}=\text{7 cm.}\\ \text{Determine :}\\ \text{(i) sin A, cos A}\\ \text{(ii) sin C, cos C}\end{array}$

Ans.

$\begin{array}{l}\text{Using Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ \text{AC\hspace{0.17em}}=\sqrt{{24}^{2}+{7}^{2}}=\sqrt{576+49}=\sqrt{625}=25\text{cm}\end{array}$

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \text{(ii)}\\ \text{sin\hspace{0.17em}C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}\\ \mathrm{cos}\text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}\end{array}$

Q.2

$\text{In the following figure},\text{}\mathrm{tanP}-\mathrm{cotR}.$

Ans.

$\begin{array}{l}\text{Using Pythagoras theorem in}\mathrm{\Delta }\text{PQR, we get}\\ \text{QR\hspace{0.17em}}=\sqrt{{13}^{2}-{12}^{2}}=\sqrt{169-144}=\sqrt{25}=5\text{cm}\end{array}$

$\begin{array}{l}\mathrm{tanP}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \mathrm{cotR}=\frac{\text{side adjacent to angle R}}{\text{side opposite to angle R}}=\frac{\text{QR}}{\text{PQ}}=\frac{5}{12}\\ \\ \mathrm{tanP}-\mathrm{cotR}=\frac{5}{12}-\frac{5}{12}=0\end{array}$

Q.3

$\text{If sin A}=\frac{3}{4},\text{calculate}\mathrm{cos}\text{A and}\mathrm{tan}\text{A.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sin A}=\frac{3}{4}\\ \text{or}\frac{\text{BC}}{\text{AC}}=\frac{3}{4}\\ \text{Let BC be 3k. Therefore, AC will be 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or (4k)}}^{2}={\text{AB}}^{2}+{\left(3\mathrm{k}\right)}^{2}\\ \text{or}16{\mathrm{k}}^{2}-9{\mathrm{k}}^{2}={\text{AB}}^{2}\\ \text{or \hspace{0.17em}}7{\mathrm{k}}^{2}={\text{AB}}^{2}\\ \text{or \hspace{0.17em} AB}=\sqrt{7}\mathrm{k}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{7}\mathrm{k}}{4\mathrm{k}}=\frac{\sqrt{7}}{4}\\ \mathrm{tan}\text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{\sqrt{7}\mathrm{k}}=\frac{3}{\sqrt{7}}\end{array}$

Q.4

$\text{Given 15 cot A}=\text{8, find sin A and sec A.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right-angled triangle right angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{15 cot A}=\text{8}\\ \text{or cot A}=\frac{8}{15}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{8}{15}\\ \text{Let AB be 8k. Therefore, BC will be 15k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{15k}\right)}^{2}+{\left(8\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=225{\mathrm{k}}^{2}+64{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=\text{289}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=17\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15\mathrm{k}}{17\mathrm{k}}=\frac{15}{17}\\ \mathrm{sec}\text{A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{17\mathrm{k}}{8\mathrm{k}}=\frac{17}{8}\end{array}$

Q.5

$\text{Given sec}\mathrm{\theta }=\frac{13}{12}\text{, calculate all other trigonometric ratios.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{sec}\mathrm{\theta }=\frac{13}{12}\\ \text{or}\frac{\text{AC}}{\text{AB}}=\frac{13}{12}\\ \text{Let AC be 13k. Therefore, AB is 12k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{BC}}^{2}={\text{AC}}^{2}-{\text{AB}}^{2}\\ {\text{or BC}}^{2}={\left(\text{13k}\right)}^{2}-{\left(12\mathrm{k}\right)}^{2}\\ {\text{or BC}}^{2}=169{\mathrm{k}}^{2}-144{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}BC}}^{2}=\text{25}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}BC}=5\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5\mathrm{k}}{13\mathrm{k}}=\frac{5}{13}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12\mathrm{k}}{13\mathrm{k}}=\frac{12}{13}\\ \mathrm{tan\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{side adjacent to angle}\mathrm{\theta }}=\frac{\text{BC}}{\text{AB}}=\frac{5\mathrm{k}}{12\mathrm{k}}=\frac{5}{12}\\ \mathrm{cot\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AB}}{\text{BC}}=\frac{12\mathrm{k}}{5\mathrm{k}}=\frac{12}{5}\\ \mathrm{cosec}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{hypotenuse}}{\text{side opposite to angle}\mathrm{\theta }}=\frac{\text{AC}}{\text{BC}}=\frac{13\mathrm{k}}{5\mathrm{k}}=\frac{13}{5}\end{array}$

Q.6

$\begin{array}{l}\text{If}\angle \text{A and}\angle \text{B are acute angles such that cos A}=\text{cos B,}\\ \text{then show that}\angle \text{A}=\angle \text{B.}\end{array}$

Ans.

$\text{Let us consider a}\mathrm{\Delta }\text{\hspace{0.17em}ABC in which CD}\perp \text{AB.}$

$\begin{array}{l}\text{Given that,}\\ \text{cos A}=\text{cos B}\\ \text{or}\frac{\text{AD}}{\text{AC}}=\frac{\text{BD}}{\text{BC}}\\ \text{or}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}\\ \text{Let}\frac{\text{AD}}{\text{BD}}=\frac{\text{AC}}{\text{BC}}=\mathrm{k}\text{}\\ \text{or AD}=\mathrm{k}\text{BD}...\text{(1)}\\ \text{and}\\ \text{AC}=\mathrm{k}\text{BC}...\text{(2)}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}CAD and}\mathrm{\Delta }\text{\hspace{0.17em}CBD, we get}\\ {\text{CD}}^{2}={\text{AC}}^{2}-{\text{AD}}^{2}\text{}...\text{(3)}\\ \text{and}\\ {\text{CD}}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\text{}...\text{(4)}\\ \text{From equations (3) and (4), we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^{2}-{\text{AD}}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{(}\mathrm{k}{\text{BC)}}^{2}-{\left(\mathrm{k}\text{BD}\right)}^{2}={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{or}{\mathrm{k}}^{2}{\text{(BC}}^{2}-{\text{BD}}^{2}\right)={\text{BC}}^{2}-{\text{BD}}^{2}\\ \text{or}{\mathrm{k}}^{2}=1\\ \text{or \hspace{0.17em}}\mathrm{k}=1\\ \text{On putting this value of}\mathrm{k}\text{in equation (2), we get}\\ \text{AC}=\text{BC}\\ \text{i.e.,}\angle \text{A}=\angle \text{B}\end{array}$

Q.7

$\text{If cot}\mathrm{\theta }=\frac{\text{7}}{8},\text{evaluate : (i)}\frac{\left(1+\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)},{\text{(ii) cot}}^{2}\mathrm{\theta }$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot}\mathrm{\theta }=\frac{7}{8}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{7}{8}\\ \text{Let BC be 8k. Therefore, AB is 7k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{7k}\right)}^{2}+{\left(8\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=49{\mathrm{k}}^{2}+64{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=11\text{3}{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=\sqrt{113}\mathrm{k}\\ \mathrm{sin\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{8\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{8}{\sqrt{113}}\\ \mathrm{cos\theta }=\frac{\text{side adjacent to angle}\mathrm{\theta }}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{7\mathrm{k}}{\sqrt{113}\mathrm{k}}=\frac{7}{\sqrt{113}}\\ \text{(i)}\frac{\left(1+\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}=\frac{1-{\mathrm{sin}}^{2}\mathrm{\theta }}{1-{\mathrm{cos}}^{2}\mathrm{\theta }}=\frac{1-{\left(\frac{8}{\sqrt{113}}\right)}^{2}}{1-{\left(\frac{7}{\sqrt{113}}\right)}^{2}}=\frac{113-64}{113-49}=\frac{49}{64}\\ {\text{(ii) cot}}^{2}\mathrm{\theta }={\left(\frac{7}{8}\right)}^{2}=\frac{49}{64}\end{array}$

Q.8

$\text{If 3 cot A}=\text{4, check whether}\frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}={\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A or not.}$

Ans.

$\text{Let}\mathrm{\Delta }\text{\hspace{0.17em}ABC is a right triangle, right-angled at B.}$

$\begin{array}{l}\text{Given that,}\\ \text{cot A}=\frac{4}{3}\\ \text{or}\frac{\text{AB}}{\text{BC}}=\frac{4}{3}\\ \text{Let BC be 3k. Therefore, AB = 4k, where k is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\text{4k}\right)}^{2}+{\left(3\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=16{\mathrm{k}}^{2}+9{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=25{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=5\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{3\mathrm{k}}{5\mathrm{k}}=\frac{3}{5}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{4\mathrm{k}}{5\mathrm{k}}=\frac{4}{5}\\ \mathrm{tan}\text{A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{\text{BC}}{\text{AB}}=\frac{3\mathrm{k}}{4\mathrm{k}}=\frac{3}{4}\\ \text{Now,}\\ \frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}=\frac{1-{\left(\frac{3}{4}\right)}^{2}}{1+{\left(\frac{3}{4}\right)}^{2}}=\frac{7}{25}\\ {\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A}={\left(\frac{4}{5}\right)}^{2}-{\left(\frac{3}{5}\right)}^{2}=\frac{7}{25}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1-{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{tan}}^{2}\text{A}}={\mathrm{cos}}^{2}\text{A}-{\mathrm{sin}}^{2}\text{A}\end{array}$

Q.9

$\begin{array}{l}\text{In triangle ABC, right-angled at B, if tan A}=\frac{1}{\sqrt{3}},\text{find the value of:}\\ \text{(i) sin A cos C}+\text{cos A sin C}\\ \text{(ii) cos A cos C}-\text{sin A sin C}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{tan A}=\frac{1}{\sqrt{3}}\\ \text{or}\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{3}}\\ \text{Let BC be}\mathrm{k}\text{. Then, AB =}\sqrt{3}\mathrm{k}\text{, where}\mathrm{k}\text{is}\\ \text{a positive integer.}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ {\text{or AC}}^{2}={\left(\sqrt{3}\text{k}\right)}^{2}+{\left(\mathrm{k}\right)}^{2}\\ {\text{or AC}}^{2}=3{\mathrm{k}}^{2}+{\mathrm{k}}^{2}\\ {\text{or \hspace{0.17em}AC}}^{2}=4{\mathrm{k}}^{2}\\ \text{or \hspace{0.17em}AC}=2\mathrm{k}\\ \mathrm{sin}\text{A}=\frac{\text{side opposite to angle A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \mathrm{cos}\text{A}=\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \mathrm{sin}\text{C}=\frac{\text{side opposite to angle C}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\mathrm{k}}{2\mathrm{k}}=\frac{\sqrt{3}}{2}\\ \mathrm{cos}\text{C}=\frac{\text{side adjacent to angle C}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\mathrm{k}}{2\mathrm{k}}=\frac{1}{2}\\ \text{Now,}\\ \text{(i)}\mathrm{sin}\text{A}\mathrm{cos}\text{C}+\mathrm{cos}\text{A}\mathrm{sin}\text{C}=\frac{1}{2}×\frac{1}{2}+\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}=1\\ \text{(ii)}\mathrm{cos}\text{A}\mathrm{cos}\text{C}-\mathrm{sin}\text{A}\mathrm{sin}\text{C}=\frac{\sqrt{3}}{2}×\frac{1}{2}-\frac{1}{2}×\frac{\sqrt{3}}{2}=0\end{array}$

Q.10

$\begin{array}{l}\text{Inâ€„}\mathrm{\Delta }\text{\hspace{0.17em}PQR, right-angled at Q, PR}+\text{QR}=\text{25 cm and PQ}=\text{5 cm.}\\ \text{Determine the values of sin\hspace{0.17em}P, cos\hspace{0.17em}P and tan\hspace{0.17em}\hspace{0.17em}P.}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{PR}+\text{QR}=\text{25 cm}\\ \text{and}\\ \text{PQ}=\text{5 cm}\\ \text{Let PR be}\mathrm{x}\text{. Then, QR =}25-\mathrm{x}\\ \text{Applying Pythagoras Theorem in}\mathrm{\Delta }\text{\hspace{0.17em}PQR, we get}\\ {\text{PR}}^{2}={\text{PQ}}^{2}+{\text{QR}}^{2}\\ \text{or}{\mathrm{x}}^{2}={\left(5\right)}^{2}+{\left(25-\mathrm{x}\right)}^{2}\\ \text{or}{\mathrm{x}}^{2}=25+625+{\mathrm{x}}^{2}-50\mathrm{x}\\ \text{or \hspace{0.17em}}50\mathrm{x}=650\\ \text{or \hspace{0.17em}}\mathrm{x}=13\\ \therefore \text{PR}=\mathrm{x}=13\text{cm}\\ \text{and}\\ \text{QR}=25-\mathrm{x}=25-13=12\text{cm}\\ \mathrm{sin}\text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{hypotenuse}}=\frac{\text{QR}}{\text{PR}}=\frac{12}{13}\\ \mathrm{cos}\text{\hspace{0.17em}P}=\frac{\text{side adjacent to angle P}}{\text{hypotenuse}}=\frac{\text{PQ}}{\text{PR}}=\frac{5}{13}\\ \mathrm{tan}\text{\hspace{0.17em}P}=\frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}}=\frac{\text{QR}}{\text{PQ}}=\frac{12}{5}\end{array}$

Q.11

$\begin{array}{l}\text{State whether the following are true or false. Justify}\\ \text{your answer.}\\ \text{(i) The value of tan A is always less than 1.}\\ \text{(ii) sec A}=\frac{12}{5}\text{â€„for some value of angle A.}\\ \text{(iii) cos A is the abbreviation used for the cosecant}\\ \text{of angle A.}\\ \text{(iv) cot A is the product of cot and A.}\\ \text{(v) sinâ€„}\mathrm{\theta }=\frac{4}{3}\text{â€„for some angleâ€„}\mathrm{\theta }.\end{array}$

Ans.

(i) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\text{}\\ \mathrm{tan}\text{\hspace{0.17em}A}=\frac{\text{side opposite to angle A}}{\text{side adjacent to angle A}}=\frac{12}{5}>1\\ \therefore \mathrm{tan}\text{\hspace{0.17em}A}>1\\ \text{So, tan\hspace{0.17em}A<1 is not always true.}\\ \text{Hence, the given statement is false.}\end{array}$

(ii) Let us consider the following Δ ABC, right-angled at B.

$\begin{array}{l}\text{We consider on the above}\mathrm{\Delta }\text{\hspace{0.17em}ABC, right-angled at B.}\\ \\ \mathrm{sec}\text{\hspace{0.17em}A}=\frac{\text{hypotenuse}}{\text{side adjacent to angle A}}=\frac{\text{AC}}{\text{AB}}=\frac{12}{5}\\ \text{Now, by using Pythagoras theorem, we have}\\ {\text{BC}}^{2}={\text{AC}}^{2}-{\text{AB}}^{2}=144-25=119\\ \text{or BC}=\sqrt{119}=10.9\\ \text{Therefore,}\mathrm{sec}\text{\hspace{0.17em}A}=\frac{12}{5}\text{is possible for some values}\\ \text{of angle A.}\\ \text{Hence, the given statement is true.}\\ \text{(iii)}\\ \text{The abbreviation used for the cosecant of angle A is cosec\hspace{0.17em}A.}\\ \text{Therefore, the given statement is false.}\\ \text{(iv)}\\ \text{cot A is not the product of cot and A. Therefore, the given}\\ \text{statement is true.}\\ \text{(v)}\\ \mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}\\ \text{In a right-angled triangle hypotenuse is the longest side.}\\ \text{Thus,}\\ \mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{\text{side opposite to angle}\mathrm{\theta }}{\text{hypotenuse}}<1\\ \text{Therefore,}\mathrm{sin}\text{\hspace{0.17em}}\mathrm{\theta }=\frac{4}{3}>1\text{is not possible for any value of}\mathrm{\theta }.\\ \text{Hence, the given statement is false.}\end{array}$

Q.12

$\begin{array}{l}\text{Evaluate the following:}\\ \text{(i) sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}\\ {\text{(ii) 2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}\\ \text{(iii)}\frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}\\ \text{(iv)}\frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\frac{{\text{5cos}}^{2}\text{60°}+{\text{4sec}}^{2}\text{30°}-{\text{tan}}^{2}\text{45°}}{{\text{sin}}^{2}\text{30°}+{\text{cos}}^{2}\text{30°}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{sin\hspace{0.17em}60°\hspace{0.17em}cos 30°}+\text{sin 30°\hspace{0.17em}cos 60°}=\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}=1\\ \text{(ii)}\\ {\text{2tan}}^{\text{2}}\text{45°}+{\text{cos}}^{\text{2}}\text{30°}-{\text{sin}}^{\text{2}}\text{60°}=2×{\left(1\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}-{\left(\frac{\sqrt{3}}{2}\right)}^{2}=2\\ \text{(iii)}\\ \frac{\text{cos 45°}}{\text{sec 30°}+\text{cosec 30°}}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\sqrt{3}}{2\sqrt{2}\left(1+\sqrt{3\right)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}}{2\sqrt{2}\left(1+\sqrt{3\right)}}×\frac{\left(1-\sqrt{3\right)}}{\left(1-\sqrt{3\right)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}-3}{2\sqrt{2}\left(1-3\right)}=\frac{3-\sqrt{3}}{4\sqrt{2}}\\ \text{(iv)}\\ \frac{\text{sin 30°}+\text{tan 45°}-\text{cosec 60°}}{\text{sec 30°}+\text{cos 60°}+\text{cot 45°}}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\\ \text{}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}×\frac{3\sqrt{3}-4}{3\sqrt{3}-4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(3\sqrt{3}-4\right)}^{2}}{{\left(3\sqrt{3}\right)}^{2}-16}=\frac{{\left(3\sqrt{3}\right)}^{2}-24\sqrt{3}+16}{27-16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{43-24\sqrt{3}}{11}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}}\\ \frac{{\text{5cos}}^{2}\text{60°}+{\text{4sec}}^{2}\text{30°}-{\text{tan}}^{2}\text{45°}}{{\text{sin}}^{2}\text{30°}+{\text{cos}}^{2}\text{30°}}=\frac{5×{\left(\frac{1}{2}\right)}^{2}+4×{\left(\frac{2}{\sqrt{3}}\right)}^{2}-1}{{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\end{array}$

Q.13

$\begin{array}{l}\text{Choose the correct option and justify your choice:}\\ \text{(i)}\frac{\text{2tan30°}}{\text{1}+{\text{tan}}^{2}\text{30°}}=\\ \text{(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°}\\ \text{(ii)}\frac{\text{1}-{\text{tan}}^{2}\text{45°}}{\text{1}+{\text{tan}}^{2}\text{45°}}=\\ \text{(A) tan 90° (B) 1 (C) sin 45° (D) 0}\\ \text{(iii) sin2A}=\text{2sinA is true when A}=\\ \text{(A) 0° (B) 30° (C) 45° (D) 60°}\\ \text{(iv)}\frac{\text{2tan30°}}{1-{\text{tan}}^{2}\text{30°}}=\\ \text{(A) cos 60° (B)sin 60° (C) tan 60° (D) sin 30}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \frac{\text{2tan30°}}{\text{1}+{\text{tan}}^{2}\text{30°}}=\frac{2×\frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}=\frac{2×3}{4\sqrt{3}}=\frac{\sqrt{3}}{2}=\mathrm{sin}60\mathrm{°}\\ \text{Hence, the correct option is (A).}\\ \text{(ii)}\\ \text{}\frac{\text{1}-{\text{tan}}^{2}\text{45°}}{\text{1}+{\text{tan}}^{2}\text{45°}}=\frac{1-1}{1+1}=0\\ \text{Hence, the correct option is (D).}\\ \text{(iii)}\\ \text{sin2A}=\text{2sinA}\\ \text{On putting A}=0\mathrm{°},\text{we get}\\ \text{sin\hspace{0.17em}}0\mathrm{°}=\text{2sin\hspace{0.17em}}0\mathrm{°}\\ \text{or 0}=2×0\\ \text{or 0}=0\\ \text{Therefore, the correct option is (A).}\\ \text{Other given options does not satisfy the given condition.}\\ \text{(iv)}\\ \frac{\text{2tan30°}}{1-{\text{tan}}^{2}\text{30°}}=\frac{\text{2}×\frac{1}{\sqrt{3}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^{2}}=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}=\frac{2}{\sqrt{3}}×\frac{3}{2}=\sqrt{3}=\mathrm{tan}\text{\hspace{0.17em}}60\mathrm{°}\\ \text{Therefore, the correct option is (C).}\end{array}$

Q.14

$\text{If tan}\left(\mathrm{A}+\mathrm{B}\right)=\sqrt{\text{3}}\text{â€„and tan}\left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{\text{3}}}\text{; 0° < A + B}\le \text{90°; A > B, find A and B.}$

Ans.

$\begin{array}{l}\text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan}\left(\mathrm{A}+\mathrm{B}\right)=\sqrt{\text{3}}\\ \text{or tan (A}+\text{B}\right)=\text{tan\hspace{0.17em}60°}\\ \text{or A}+\text{B}=60\mathrm{°}\text{}...\text{(1)}\\ \text{Again, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan}\left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{\text{3}}}\\ \text{or tan (A}-\text{B}\right)=\text{tan\hspace{0.17em}}3\text{0°}\\ \text{or A}-\text{B}=30\mathrm{°}\text{}...\text{(2)}\\ \text{On adding equations (1) and (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 2A}=90\mathrm{°}\\ \text{or A}=45\mathrm{°}\\ \text{On putting this value of A in equation (1), we get}\\ \text{B}=15\mathrm{°}\end{array}$

Q.15

$\begin{array}{l}\text{State whether the following are true or false.}\\ \text{Justify your answer.}\\ \text{(i) sin}\left(\mathrm{A}+\mathrm{B}\right)=\text{sin A}+\text{sin B.}\\ \text{(ii) The value of sin}\mathrm{\theta }\text{â€„increases asâ€„}\mathrm{\theta }\text{â€„increases.}\\ \text{(iii) The value of cos}\mathrm{\theta }\text{â€„increases asâ€„}\mathrm{\theta }\text{â€„increases.}\\ \text{(iv) sin}\mathrm{\theta }=\text{cos}\mathrm{\theta }\text{â€„for all values ofâ€„}\mathrm{\theta }.\\ \text{(v) cot A is not defined for A}=0.\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let A}=30\mathrm{°}\text{and B}=60\mathrm{°}.\\ \text{Now,}\\ \mathrm{sin}\left(\text{A}+\text{B)}=\mathrm{sin}\left(30\mathrm{°}+60\mathrm{°}\text{)}=\mathrm{sin}90\mathrm{°}=1\\ \text{and}\\ \text{sin A}+\text{sin B}=\text{sin}30\mathrm{°}+\text{sin\hspace{0.17em}\hspace{0.17em}}60\mathrm{°}=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\\ \therefore \mathrm{sin}\left(\text{A}+\text{B)}\ne \text{sin A}+\text{sin B}\\ \text{Hence, the given statement is false.}\\ \text{(ii)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{°}=0,\\ \mathrm{sin}30\mathrm{°}=\frac{1}{2}=0.5,\\ \mathrm{sin}45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{sin}60\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{sin}90\mathrm{°}=1\\ \text{We see that the value of sin}\mathrm{\theta }\text{increases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is true if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iii)}\\ \text{We know that}\\ \mathrm{cos}0\mathrm{°}=1,\\ \mathrm{cos}30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87,\\ \mathrm{cos}45\mathrm{°}=\frac{1}{\sqrt{2}}=0.7,\\ \mathrm{cos}60\mathrm{°}=\frac{1}{2}=0.5,\\ \mathrm{cos}90\mathrm{°}=0\\ \text{We see that the value of}\mathrm{cos\theta }\text{decreases as}\mathrm{\theta }\text{increases}\\ \text{in the interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{Hence, the given statement is false if}\mathrm{\theta }\text{lies in the}\\ \text{interval}0\mathrm{°}\le \mathrm{\theta }\le 90\mathrm{°}.\\ \text{(iv)}\\ \text{We know that}\\ \mathrm{sin}0\mathrm{°}=0\text{and}\mathrm{cos}0\mathrm{°}=1,\\ \mathrm{sin}30\mathrm{°}=\frac{1}{2}=0.5\text{and}\mathrm{cos}30\mathrm{°}=\frac{\sqrt{3}}{2}=0.87.\\ \text{Therefore, sin}\mathrm{\theta }\ne \text{cos}\mathrm{\theta }\text{for all values of}\mathrm{\theta }.\\ \text{Hence, the given statement is false}.\\ \text{(v)}\\ \text{We know that cot A}=\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}.\text{}\\ \text{So, for A}=0,\text{we have}\\ \text{cot 0}=\frac{\mathrm{cos}\text{0}}{\mathrm{sin}\text{0}}=\frac{1}{0}\\ \text{We know that division by 0 is not defined. Therefore, cot 0}\\ \text{is not defined.}\\ \text{Hence, the given statement is true}.\end{array}$

Q.16

$\begin{array}{l}\text{Evaluate\hspace{0.17em}:}\\ \text{(i)}\frac{\text{sin 18°}}{\text{cos 72}°}\text{}\\ \text{(ii)}\frac{\text{tan 26°}}{\text{cot 64°}}\text{}\\ \text{(iii) cos 48°}-\text{sin 42°}\\ \text{(iv) cosec 31°}-\text{sec 59°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{}\frac{\text{sin 18°}}{\text{cos 72}\mathrm{°}}=\frac{\text{cos(90°}-18\text{°}\right)}{\text{cos 72}\mathrm{°}}=\frac{\text{cos 72}\mathrm{°}}{\text{cos 72}\mathrm{°}}=1\\ \\ \text{(ii)}\\ \text{}\frac{\text{tan 26°}}{\text{cot 64°}}=\frac{\text{cot (90°}-\text{26°)}}{\text{cot 64°}}=\frac{\text{cot 64°}}{\text{cot 64°}}=1\\ \\ \text{(iii)}\\ \text{cos 48°}-\text{sin 42°}=\text{sin (90°}-\text{48°)}-\text{sin 42°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{sin 42°}-\text{sin 42°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \text{(iv)}\\ \text{cosec 31°}-\text{sec 59°}=\text{sec (90°}-\text{31°)}-\text{sec 59°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{sec 59°}-\text{sec 59°}\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\end{array}$

Q.17

$\begin{array}{l}\text{Show that\hspace{0.17em}:}\\ \text{(i) tan 48° tan 23° tan 42° tan 67°}=1\\ \text{(ii) cos 38° cos 52°}-\text{sin 38° sin 52°}=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{tan 48° tan 23° tan 42° tan 67°}\\ =\text{cot (90°}-\text{48°) cot (90°}-23\text{°) tan 42° tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em} cot 67°\hspace{0.17em}\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67°}\\ =\text{cot}42°\text{\hspace{0.17em}tan 42°\hspace{0.17em}\hspace{0.17em}tan 67° cot 67°}\\ =1×1=1\\ \text{(ii)}\\ \text{cos 38° cos 52°}-\text{sin 38° sin 52°}\\ =\mathrm{sin}\text{(90°}-3\text{8°)\hspace{0.17em}cos (90°}-3\text{8°)}-\text{sin 38° sin 52°}\\ =\mathrm{sin}\text{\hspace{0.17em}\hspace{0.17em}52°\hspace{0.17em}}\mathrm{sin}38\text{°}-\text{sin 38° sin 52°}\\ =0\end{array}$

Q.18

$\text{If tan 2A}=\text{cot}\left(\text{A}-18\mathrm{°}\right),\text{where 2A is an acute angle},\text{find the value of A}.$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{tan 2A}=\text{cot}\left(\text{A}-18\mathrm{°}\right)\\ \text{or cot (90°}-2\text{A}\right)=\text{cot}\left(\text{A}-18\mathrm{°}\right)\\ \text{or 90°}-2\text{A}=\text{A}-18\mathrm{°}\\ \text{or}3\text{A}=108\mathrm{°}\\ \text{or A}=\frac{108\mathrm{°}}{3}=36\mathrm{°}\end{array}$

Q.19

$\text{If tan A}=\text{cot B, prove that A}+\text{B}=\text{90°.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} tan A}=\text{cot B}\\ \text{or cot (90°}-\text{A)}=\mathrm{cot}\text{\hspace{0.17em}\hspace{0.17em}B}\\ \text{or 90°}-\text{A}=\text{B}\\ \text{or A}+\text{B}=\text{90°}\end{array}$

Q.20

$\text{If sec 4A}=\text{cosec (A}-\text{20°), where 4A is an acute angle, find the value of A.}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{sec 4A}=\text{cosec (A}-\text{20°)}\\ \text{or cosec (90°}-\text{4A)}=\text{cosec (A}-\text{20°)}\\ \text{or 90°}-\text{4A}=\text{A}-\text{20°}\\ \text{or 5A}=110\text{°}\\ \text{or A}=\frac{110\text{°}}{5}=22\text{°}\end{array}$

Q.21

$\begin{array}{l}\text{If A, B and C are interior angles of a triangle ABC,}\\ \text{then show that}\\ \text{sin}\left(\frac{\text{B+C}}{2}\right)=\mathrm{cos}\frac{\text{A}}{2}.\end{array}$

Ans.

$\begin{array}{l}\text{Given that A, B and C are interior angles of a triangle ABC.}\\ \therefore \text{A}+\text{B}+\text{C}=180\mathrm{°}\\ \text{or A}=180\mathrm{°}-\text{B}-\text{C}\\ \text{Now,}\\ \text{sin}\left(\frac{\text{B}+\text{C}}{2}\right)=\mathrm{cos}\left(90\mathrm{°}-\frac{\text{B}+\text{C}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cos}\left(\frac{180\mathrm{°}-\text{B}-\text{C}}{2}\right)\\ \text{}=\mathrm{cos}\left(\frac{\text{A}}{2}\right)\text{}\end{array}$

Q.22

$\text{Express sin 67°}+\text{cos 75° in terms of trigonometric ratios of angles between 0° and 45°.}$

Ans.

$\begin{array}{l}\text{sin 67°}+\text{cos 75°}=\mathrm{cos}\text{(90°}-\text{67°)}+\text{sin (90°}-\text{75°)}\\ \text{}=\mathrm{cos}23\text{°}+\text{sin 15°}\end{array}$

Q.23

$\text{Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{cosec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{cot}}^{2}\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{sin}}^{2}\text{A}}=1+{\text{cot}}^{2}\text{\hspace{0.17em}A}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\mathrm{sin}}^{2}\text{A}=\frac{1}{1+{\text{cot}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\mathrm{sin}\text{A}=\frac{1}{\sqrt{1+{\text{cot}}^{2}\text{\hspace{0.17em}A}}}\\ \text{Also, we know that}\\ {\text{sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+{\text{tan}}^{2}\text{\hspace{0.17em}A}\\ {\text{or sec}}^{\text{2}}\text{\hspace{0.17em}A}=1+\frac{1}{{\text{cot}}^{2}\text{\hspace{0.17em}A}}\\ \text{or}\mathrm{sec}\text{\hspace{0.17em}A}=\sqrt{\frac{1+{\mathrm{cot}}^{2}\text{A}}{{\mathrm{cot}}^{2}\text{A}}}=\frac{\sqrt{1+{\mathrm{cot}}^{2}\text{A}}}{\mathrm{cot}\text{A}}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A}=\frac{\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=\frac{1}{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}}=\frac{1}{\mathrm{cot}\text{A}}\end{array}$

Q.24

$\text{Write all the other trigonometric ratios of}\angle \text{A in terms of sec A.}$

Ans.

$\begin{array}{l}\text{We know that}\\ {\text{sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-{\text{cos}}^{2}\text{\hspace{0.17em}A}\\ {\text{or \hspace{0.17em}\hspace{0.17em}sin}}^{\text{2}}\text{\hspace{0.17em}A}=1-\frac{1}{{\text{sec}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}\hspace{0.17em}}{\mathrm{sin}}^{2}\text{A}=\frac{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}{{\text{sec}}^{2}\text{\hspace{0.17em}A}}\\ \text{or \hspace{0.17em}}\mathrm{sin}\text{A}=\frac{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\text{}\frac{1}{\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}=\frac{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}{\text{sec\hspace{0.17em}A}}\\ \mathrm{or}\text{}\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}=\frac{\text{sec\hspace{0.17em}A}}{\sqrt{{\text{sec}}^{2}\text{\hspace{0.17em}A}-1}}\\ \text{Also, we know that}\\ \text{sec\hspace{0.17em}A cos\hspace{0.17em}A}=1\\ \text{or cos\hspace{0.17em}A}=\frac{1}{\text{sec\hspace{0.17em}A}}\\ \\ \text{Also, we know that}\\ {\text{sec}}^{2}\text{A}-{\text{tan}}^{2}\text{\hspace{0.17em}A}=1\\ {\text{or tan}}^{2}\text{\hspace{0.17em}A}={\text{sec}}^{2}\text{A}-1\\ \mathrm{or}\text{tan\hspace{0.17em}A}=\sqrt{{\text{sec}}^{2}\text{A}-1}\\ \text{Also, we know that}\\ \text{tan\hspace{0.17em}A cot\hspace{0.17em}A}=1\\ \mathrm{or}\text{cot\hspace{0.17em}A}=\frac{1}{\text{tan\hspace{0.17em}A}}=\frac{1}{\sqrt{{\text{sec}}^{2}\text{A}-1}}\end{array}$

Q.25

$\begin{array}{l}\text{Evaluate:}\\ \text{(i)}\frac{{\text{sin}}^{2}\text{63°}+{\text{sin}}^{2}\text{27°}}{{\text{cos}}^{2}\text{17°}+{\text{cos}}^{2}\text{73°}}\\ \text{(ii) sin 25° cos 65°}+\text{cos 25° sin 65°}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{}\frac{{\text{sin}}^{2}\text{63°}+{\text{sin}}^{2}\text{27°}}{{\text{cos}}^{2}\text{17°}+{\text{cos}}^{2}\text{73°}}=\frac{{\text{sin}}^{2}\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^{2}\text{(90°}-\text{27°)}}{{\text{sin}}^{2}\text{(90°}-\text{17°)}+{\text{cos}}^{2}\text{\hspace{0.17em}73°}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\text{sin}}^{2}\text{\hspace{0.17em}63°}+\mathrm{co}{\text{s}}^{2}\text{\hspace{0.17em}63°}}{{\text{sin}}^{2}\text{\hspace{0.17em}73°}+{\text{cos}}^{2}\text{\hspace{0.17em}73°}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{1}=1\\ \text{(ii)}\\ \text{sin\hspace{0.17em}25° cos 65°}+\text{cos\hspace{0.17em}25° sin\hspace{0.17em}65°}\\ =\text{sin\hspace{0.17em}25°\hspace{0.17em}sin(90°}-\text{65°}\right)+\text{cos 25° cos(90°}-\text{\hspace{0.17em}65°)}\\ =\text{sin\hspace{0.17em}25° sin\hspace{0.17em}25°}+\text{cos 25°\hspace{0.17em}cos 25°}\\ ={\text{sin}}^{2}\text{\hspace{0.17em}25°}+{\text{cos}}^{2}\text{25°}\\ =1\end{array}$

Q.26

$\begin{array}{l}\text{Choose the correct option. Justify your choice.}\\ {\text{(i) 9 sec}}^{2}\text{A}-{\text{9 tan}}^{2}\text{A}=\\ \text{(A) 1 (B) 9 (C) 8 (D) 0}\\ \text{(ii) (1 + tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (1 + cot}\mathrm{\theta }-\text{cosec}\mathrm{\theta }\text{)}=\\ \text{(A) 0 (B) 1 (C) 2 (D)}-\text{1}\\ \text{(iii) (sec A + tan A) (1}-\text{sin A)}=\\ \text{(A) sec A (B) sin A (C) cosec A (D) cos A}\\ \text{(iv)}\frac{1+{\mathrm{tan}}^{2}\mathrm{A}}{1+{\mathrm{cot}}^{2}\text{}\mathrm{A}}=\\ {\text{(A) sec}}^{2}\mathrm{A}\text{(B)}-{\text{1 (C) cot}}^{2}{\text{A (D) tan}}^{2}\mathrm{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ {\text{9 sec}}^{2}\text{A}-{\text{9 tan}}^{2}\text{A}=9\left({\text{sec}}^{2}\text{A}-{\text{tan}}^{2}\text{A)}=9×1=9\\ \text{Therefore, the correct option is (B).}\\ \text{(ii)}\\ \text{(1}+\text{tan}\mathrm{\theta }+\text{sec}\mathrm{\theta }\text{) (}1+\mathrm{cot\theta }-\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{\theta }\text{)}\\ =\left(1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{1}{\mathrm{cos\theta }}\right)\left(1+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}-\frac{1}{\mathrm{sin\theta }}\right)\\ =\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }+1}{\mathrm{cos\theta }}\right)\left(\frac{\mathrm{sin\theta }+\mathrm{cos\theta }-1}{\mathrm{sin\theta }}\right)\\ =\frac{{\left(\mathrm{sin\theta }+\mathrm{cos\theta }\right)}^{2}-1}{\mathrm{sin\theta cos\theta }}=\frac{{\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}\\ =\frac{1+2\mathrm{sin\theta cos\theta }-1}{\mathrm{sin\theta cos\theta }}=2\\ \text{Therefore, the correct option is (C).}\\ \text{(iii)}\\ \text{(sec A + tan A) (1}-\text{sin A)}\\ =\left(\frac{1}{\text{cos A}}+\frac{\mathrm{sin}\text{A}}{\text{cos A}}\right)\left(\text{1}-\text{sin A}\right)\\ =\left(\frac{1+\mathrm{sin}\text{A}}{\text{cos A}}\right)\left(\text{1}-\text{sin A}\right)\\ =\frac{1-{\mathrm{sin}}^{2}\text{A}}{\text{cos A}}=\frac{{\mathrm{cos}}^{2}\text{A}}{\text{cos A}}=\mathrm{cos}\text{A}\\ \text{Therefore, the correct option is (D).}\\ \text{(iv)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}=\frac{{\mathrm{sec}}^{2}\text{A}}{{\mathrm{cosec}}^{2}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{{\mathrm{cos}}^{2}\text{A}}={\mathrm{tan}}^{2}\text{A}\\ \text{Therefore, the correct option is (D).}\end{array}$

Q.27

$\begin{array}{l}\text{Prove the following identities, where the angles involved}\\ \text{are acute angles for which the expressions are defined.}\\ \text{(i) (cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta }{\right)}^{2}=\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}\\ \text{(ii)}\frac{\mathrm{cos}\text{A}}{1+\mathrm{sin}\text{A}}+\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=2\mathrm{sec}\text{A}\\ \text{(iii)}\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}=1+\mathrm{sec\theta cosec\theta }\\ \text{[Hint :Write the expression in terms of sin}\mathrm{\theta }\text{and cos}\mathrm{\theta }.\right]\\ \text{(iv)}\frac{1+\mathrm{sec}\text{A}}{\mathrm{sec}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(v)}\frac{\mathrm{cos}\text{A}-\mathrm{sin}\text{A}+1}{\mathrm{cos}\text{A}+\mathrm{sin}\text{A}-1}=\text{cosec A}+\text{cot A, using the identity}\\ {\text{cosec}}^{\text{2}}\text{A}=\text{1}+{\text{cot}}^{2}\text{\hspace{0.17em}A.}\\ \text{(vi)}\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}}=\mathrm{sec}\text{A}+\mathrm{tan}\text{A}\\ \text{(vii)}\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^{3}\mathrm{\theta }}{{\text{2cos}}^{3}\mathrm{\theta }-\text{cos}\mathrm{\theta }}=\text{tan}\mathrm{\theta }\\ {\text{(viii) (sin A + cosec A)}}^{2}+{\left(\text{cos A}+\text{sec A}\right)}^{\text{2}}=7+{\mathrm{tan}}^{2}\text{A}+{\mathrm{cot}}^{2}\text{A}\\ \text{(ix) (cosec A}-\text{sin A) (sec A}-\text{cos A)}=\frac{1}{\mathrm{tan}\text{A}+\mathrm{cot}\text{A}}\\ \text{[Hint: Simplify LHS and RHS separately]}\\ \text{(x)}\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^{2}={\mathrm{tan}}^{2}\text{A}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{LHS}={\left(\text{cosec}\mathrm{\theta }-\text{cot}\mathrm{\theta }\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{cosec}}^{2}\mathrm{\theta }+{\mathrm{cot}}^{2}\mathrm{\theta }-2\text{cosec}\mathrm{\theta }\text{\hspace{0.17em}cot}\mathrm{\theta }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{{\mathrm{sin}}^{2}\mathrm{\theta }}+\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}-\frac{2\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1+{\mathrm{cos}}^{2}\mathrm{\theta }-2\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(1-\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}{1-{\mathrm{cos}}^{2}\mathrm{\theta }}=\frac{\left(1-\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}=\text{RHS}\\ \\ \text{(ii)}\\ \text{LHS}=\frac{\mathrm{cos}\text{A}}{1+\mathrm{sin}\text{A}}+\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}\\ \text{}=\frac{{\mathrm{cos}}^{2}\text{A}+{\left(1+\mathrm{sin}\text{A)}}^{2}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \text{}=\frac{{\mathrm{cos}}^{2}\text{A}+{\mathrm{sin}}^{2}\text{A}+2\mathrm{sin}\text{A}+1}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1+1+2\mathrm{sin}\text{A}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}\\ \text{}=\frac{2\left(1+\mathrm{sin}\text{A)}}{\left(1+\mathrm{sin}\text{A)}\mathrm{cos}\text{A}}=2\mathrm{sec}\text{A}=\text{RHS}\\ \text{}\\ \text{(iii)}\\ \text{LHS}=\frac{\mathrm{tan\theta }}{1-\mathrm{cot\theta }}+\frac{\mathrm{cot\theta }}{1-\mathrm{tan\theta }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{1-\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{1-\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{\mathrm{sin\theta }}}+\frac{\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}{\frac{\mathrm{cos\theta }-\mathrm{sin\theta }}{\mathrm{cos\theta }}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{2}\mathrm{\theta }}{\mathrm{cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}+\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{\mathrm{sin\theta }\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{3}\mathrm{\theta }-{\mathrm{cos}}^{3}\mathrm{\theta }}{\mathrm{sin\theta cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }+\mathrm{sin\theta cos\theta }\right)\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}{\mathrm{sin\theta cos\theta }\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}\\ \text{}=\frac{1+\mathrm{sin\theta cos\theta }}{\mathrm{sin\theta cos\theta }}=1+\mathrm{sec\theta cosec}\text{\hspace{0.17em}}\mathrm{\theta }=\text{RHS}\\ \text{(iv)}\\ \text{LHS}=\frac{1+\mathrm{sec}\text{A}}{\mathrm{sec}\text{A}}=\frac{1+\frac{1}{\mathrm{cos}\text{A}}}{\frac{1}{\mathrm{cos}\text{A}}}=\frac{1+\mathrm{cos}\text{A}}{1}×\frac{1-\mathrm{cos}\text{A}}{1-\mathrm{cos}\text{A}}\\ \text{}=\frac{1-{\mathrm{cos}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}=\frac{{\mathrm{sin}}^{2}\text{A}}{1-\mathrm{cos}\text{A}}=\text{RHS}\\ \text{(v)}\\ \text{LHS}=\frac{\mathrm{cos}\text{A}-\mathrm{sin}\text{A}+1}{\mathrm{cos}\text{A}+\mathrm{sin}\text{A}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}-\frac{\mathrm{sin}\text{A}}{\mathrm{sin}\text{A}}+\frac{1}{\mathrm{sin}\text{A}}}{\frac{\mathrm{cos}\text{A}}{\mathrm{sin}\text{A}}+\frac{\mathrm{sin}\text{A}}{\mathrm{sin}\text{A}}-\frac{1}{\mathrm{sin}\text{A}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cot}\text{A}-1+\mathrm{cosec}\text{A}}{\mathrm{cot}\text{A}+1-\mathrm{cosec}\text{\hspace{0.17em}A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{A)}}{\mathrm{cot}\text{A}+\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}×\frac{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{\mathrm{cot}\text{A}-\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{cot}}^{2}\text{A}+{\left(1-\mathrm{cosec}\text{A)}}^{2}-2\mathrm{cotA}\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}{{\mathrm{cot}}^{2}\text{A}-{\left(1-\mathrm{cosec}\text{\hspace{0.17em}A)}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{cot}}^{2}\text{A}+1+{\mathrm{cosec}}^{2}\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{{\mathrm{cot}}^{2}\text{A}-1-{\mathrm{cosec}}^{2}\text{\hspace{0.17em}A}+2\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{cosec}}^{2}\text{A}-\text{2}\mathrm{cosec}\text{A}-2\mathrm{cotA}+2\mathrm{cotAcosecA}}{-1-1+2\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cosec}\text{A(}\mathrm{cosec}\text{A}-\text{1}\right)+\mathrm{cotA}\left(\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{A}-1\right)}{\mathrm{cosec}\text{\hspace{0.17em}A}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cosec}\text{A}+\text{cot\hspace{0.17em}A}\\ \text{(vi)}\\ \text{LHS}=\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}}=\sqrt{\frac{1+\mathrm{sin}\text{A}}{1-\mathrm{sin}\text{A}}×\frac{1+\mathrm{sin}\text{A}}{1+\mathrm{sin}\text{A}}}\\ \text{}=\left(1+\mathrm{sin}\text{A}\right)\sqrt{\frac{1}{1-{\mathrm{sin}}^{2}\text{A}}}=\frac{1+\mathrm{sin}\text{A}}{\mathrm{cos}\text{A}}=\mathrm{tanA}+\mathrm{secA}=\text{RHS}\\ \text{(vii)}\\ \text{LHS}=\frac{\text{sin}\mathrm{\theta }-{\text{2sin}}^{3}\mathrm{\theta }}{{\text{2cos}}^{3}\mathrm{\theta }-\text{cos}\mathrm{\theta }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left({\text{2cos}}^{2}\mathrm{\theta }-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(\text{2(1}-{\mathrm{sin}}^{2}\mathrm{\theta }\right)-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(\text{2}-2{\mathrm{sin}}^{2}\mathrm{\theta }-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{sin}\mathrm{\theta }\left(1-{\text{2sin}}^{2}\mathrm{\theta }\right)}{\text{cos}\mathrm{\theta }\left(1-2{\mathrm{sin}}^{2}\mathrm{\theta }\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tan\theta }=\mathrm{RHS}\\ \text{(viii)}\\ \text{LHS}={\left(\text{sin A}+\text{cosec A}\right)}^{2}+{\left(\text{cos A}+\text{sec A}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cosec}}^{2}\mathrm{A}+2+{\mathrm{cos}}^{2}\mathrm{A}+{\mathrm{sec}}^{2}\mathrm{A}+2\\ \text{}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}+4+{\mathrm{sec}}^{2}\mathrm{A}+{\mathrm{cosec}}^{2}\mathrm{A}\\ \text{}=1+4+1+{\mathrm{tan}}^{2}\mathrm{A}+1+{\mathrm{cot}}^{2}\mathrm{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7+{\mathrm{tan}}^{2}\mathrm{A}+{\mathrm{cot}}^{2}\mathrm{A}=\text{RHS}\\ \text{(ix)}\\ \text{LHS}=\text{(cosec A}-\text{sin A) (sec A}-\text{cos A)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1}{\mathrm{sinA}}-\mathrm{sinA}\right)\left(\frac{1}{\mathrm{cosA}}-\mathrm{cosA}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1-{\mathrm{sin}}^{2}\mathrm{A}}{\mathrm{sinA}}\right)\left(\frac{1-{\mathrm{cos}}^{2}\mathrm{A}}{\mathrm{cosA}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{cos}}^{2}\mathrm{A}}{\mathrm{sinA}}×\frac{{\mathrm{sin}}^{2}\mathrm{A}}{\mathrm{cosA}}=\frac{\mathrm{sinAcosA}}{1}=\frac{\mathrm{sinAcosA}}{{\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\frac{{\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}}{\mathrm{sinAcosA}}}=\frac{1}{\frac{\mathrm{sinA}}{\mathrm{cosA}}+\frac{\mathrm{cosA}}{\mathrm{sinA}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{tanA}+\mathrm{cotA}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{RHS}\\ \text{(x)}\\ \text{LHS}=\frac{1+{\mathrm{tan}}^{2}\text{A}}{1+{\mathrm{cot}}^{2}\text{A}}=\frac{{\mathrm{sec}}^{2}\mathrm{A}}{{\mathrm{cosec}}^{2}\mathrm{A}}=\frac{\frac{1}{{\mathrm{cos}}^{2}\mathrm{A}}}{\frac{1}{{\mathrm{sin}}^{2}\mathrm{A}}}=\frac{{\mathrm{sin}}^{2}\mathrm{A}}{{\mathrm{cos}}^{2}\mathrm{A}}={\mathrm{tan}}^{2}\mathrm{A}=\mathrm{RHS}\\ \text{LHS}={\left(\frac{\text{1}-\text{tanA}}{\text{1}-\text{cot A}}\right)}^{2}={\left(\frac{1-\frac{\mathrm{sinA}}{\mathrm{cosA}}}{1-\frac{\mathrm{cosA}}{\mathrm{sinA}}}\right)}^{2}={\left(\frac{\frac{\mathrm{cosA}-\mathrm{sinA}}{\mathrm{cosA}}}{\frac{\mathrm{sinA}-\mathrm{cosA}}{\mathrm{sinA}}}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{2}\mathrm{A}}{{\mathrm{cos}}^{2}\mathrm{A}}={\mathrm{tan}}^{2}\mathrm{A}=\mathrm{RHS}\end{array}$

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### 1. What are Congruent Triangles?

Ans. Two triangles are said to be congruent to each other if the three sides and the three angles of the triangles are equal in some orientation. They superimpose each other. In other words, we can say that their shape and dimensions are the same. The symbol ≅ is used to indicate the congruence between them.

### 2. What does CPCT stand for?

CPCT stands for Corresponding Parts of Congruent triangles. According to CPCT, if two or more triangles which are congruent to each other are taken, then their corresponding angles and the sides will also be congruent to each other.

### 3. What are the different rules of congruency of triangles?

There are 5 main rules of congruence:

SSS rule: Side-Side-Side

SAS rule: Side-Angle-Side

ASA rule: Angle-Side-Angle

AAS rule: Angle-Angle-Side

RHS rule: Right angle- Hypotenuse-Side

### 4. Why should I opt to study at Extramarks?

Extramarks has developed comprehensive study material for all subjects of the NCERT syllabus. The experienced subject experts  at Extramarks  guide the students in resolving  their doubts and help them in better and clear understanding of the concepts. They help you to build  your foundation and boost confidence in Mathematics. Since the teachers understand the student psychology and interact accordingly to resolve their issues, learning at  Extramarks is fun and will eventually lead to significant improvement in your overall performance.

### 5. Explain the properties of angles.

The  different properties of angles are:

• The vertically opposite angles are always equal.
• Two adjacent angles are said to form a linear pair only if their sum is 180 degrees.
• The alternate angles and corresponding angles are equal when two parallel lines are intersected by a transversal.
• The sum of interior opposite angles on the same side of the transversal is always 180 degrees.
• In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of its other two sides.

### 6. What will the students learn after going through the Extramarks solutions of Chapter 7 of Class 7 Mathematics?

The usefulness of the solutions after going through the fundamental ideas of congruence outlined in the chapter will be appreciated by all students. They will become confident and will be able to solve any question related to this chapter. You may even try applying the congruence requirements for line segments, angles, and triangles to different plane  figures after going through  this chapter. Extramarks has outlined all the concepts of Chapter 7 Class 7 Mathematics in the solutions. You can access them for class assignments, tests and exam preparation from our website.