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

(Include NCERT Solutions for Class 7 Mathematics Chapter 7 Congruence of Triangles)

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

AC = XZ [Adjacent sides]

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

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NCERT Solutions for Class 7

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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. 

Q.1

In Δ ABC, right-angled at B, AB=24 cm, BC=7 cm.Determine :(i) sin A, cos A(ii) sin C, cos C

Ans.

Using Pythagoras Theorem in Δ ABC, we getAC =242+72=576+49=625=25 cm

(i) sin A=side opposite to angle Ahypotenuse=BCAC=725cosA=side adjacent to angle Ahypotenuse=ABAC=2425(ii) sin C=side opposite to angle Chypotenuse=ABAC=2425cosC=side adjacent to angle Chypotenuse=BCAC=725

Q.2

In the following figure, tanPcotR.

Ans.

Using Pythagoras theorem in Δ PQR, we getQR =132122=169144=25=5 cm

tanP=side opposite to angle Pside adjacent to angle P=QRPQ=512cotR=side adjacent to angle Rside opposite to angle R=QRPQ=512tanPcotR=512512=0

Q.3

If sin A=34, calculate cosA and tanA.

Ans.

Let Δ ABC is a right-angled triangle right angled at B.

Given that, sin A=34or BCAC=34Let BC be 3k. Therefore, AC will be 4k, where k isa positive integer.Applying Pythagoras Theorem in ΔABC, we get AC2=AB2+BC2or (4k)2=AB2+(3k)2or 16k29k2=AB2or   7k2=AB2or   AB=7kcosA=side adjacent to angle Ahypotenuse=ABAC=7k4k=74tanA=side opposite to angle Aside adjacent to angle A=BCAB=3k7k=37

Q.4

Given 15 cot A=8, find sin A and sec A.

Ans.

Let Δ ABC is a right-angled triangle right angled at B.

Given that, 15 cot A=8or cot A=815or ABBC=815Let AB be 8k. Therefore, BC will be 15k, where k is a positive integer.Applying Pythagoras theorem in ΔABC, we get AC2=AB2+BC2or AC2=(15k)2+(8k)2or AC2=225k2+64k2or  AC2=289k2or  AC=17ksinA=side opposite to angle Ahypotenuse=BCAC=15k17k=1517secA=hypotenuseside adjacent to angle A=ACAB=17k8k=178

Q.5

Given sec θ=1312, calculate all other trigonometric ratios.

Ans.

Let Δ ABC is a right triangle, right-angled at B.

Given that, sec θ=1312or ACAB=1312Let AC be 13k. Therefore, AB is 12k, where k isa positive integer.Applying Pythagoras Theorem in ΔABC, we get AC2=AB2+BC2 BC2=AC2AB2or BC2=(13k)2(12k)2or BC2=169k2144k2or  BC2=25k2or  BC=5ksinθ=side opposite to angle θhypotenuse=BCAC=5k13k=513cosθ=side adjacent to angle θhypotenuse=ABAC=12k13k=1213tanθ=side opposite to angle θside adjacent to angle θ=BCAB=5k12k=512cotθ=side adjacent to angle θside opposite to angle θ=ABBC=12k5k=125cosecθ=hypotenuseside opposite to angle θ=ACBC=13k5k=135

Q.6

If A and B are acute angles such that cos A=cos B,then show that A=B.

Ans.

Let us consider a Δ ABC in which CDAB.

Given that, cos A=cos Bor ADAC=BDBCor ADBD=ACBCLet ADBD=ACBC=k or AD=kBD ...(1)and AC=kBC ...(2)Applying Pythagoras Theorem in Δ CAD and Δ CBD, we get CD2=AC2AD2 ...(3)and CD2=BC2BD2 ...(4)From equations (3) and (4), we get        AC2AD2=BC2BD2 (kBC)2(kBD)2=BC2BD2or k2(BC2BD2)=BC2BD2or k2=1or  k=1On putting this value of k in equation (2), we get AC=BCi.e., A=B

Q.7

If cot θ=78, evaluate : (i) (1+sinθ)(1sinθ)(1+cosθ)(1cosθ), (ii) cot2θ

Ans.

Let Δ ABC is a right triangle, right-angled at B.

Given that, cot θ=78or ABBC=78Let BC be 8k. Therefore, AB is 7k, where k isa positive integer.Applying Pythagoras Theorem in ΔABC, we get AC2=AB2+BC2or AC2=(7k)2+(8k)2or AC2=49k2+64k2or  AC2=113k2or  AC=113ksinθ=side opposite to angle θhypotenuse=BCAC=8k113k=8113cosθ=side adjacent to angle θhypotenuse=ABAC=7k113k=7113(i) (1+sinθ)(1sinθ)(1+cosθ)(1cosθ)=1sin2θ1cos2θ=1(8113)21(7113)2=1136411349=4964(ii) cot2θ=(78)2=4964

Q.8

If 3 cot A=4, check whether 1tan2A1+tan2A=cos2Asin2A or not.

Ans.

Let Δ ABC is a right triangle, right-angled at B.

Given that, cot A=43or ABBC=43Let BC be 3k. Therefore, AB = 4k, where k isa positive integer.Applying Pythagoras theorem in ΔABC, we get AC2=AB2+BC2or AC2=(4k)2+(3k)2or AC2=16k2+9k2or  AC2=25k2or  AC=5ksinA=side opposite to angle Ahypotenuse=BCAC=3k5k=35cosA=side adjacent to angle Ahypotenuse=ABAC=4k5k=45tanA=side opposite to angle Aside adjacent to angle A=BCAB=3k4k=34Now,1tan2A1+tan2A=1(34)21+(34)2=725cos2Asin2A=(45)2(35)2=725   1tan2A1+tan2A=cos2Asin2A

Q.9

In triangle ABC, right-angled at B, if tan A=13,find the value of: (i) sin A cos C+cos A sin C (ii) cos A cos Csin A sin C

Ans.

Given that, tan A=13or BCAB=13Let BC be k. Then, AB = 3k, where k isa positive integer.Applying Pythagoras Theorem in ΔABC, we get AC2=AB2+BC2or AC2=(3k)2+(k)2or AC2=3k2+k2or  AC2=4k2or  AC=2ksinA=side opposite to angle Ahypotenuse=BCAC=k2k=12cosA=side adjacent to angle Ahypotenuse=ABAC=3k2k=32sinC=side opposite to angle Chypotenuse=ABAC=3k2k=32cosC=side adjacent to angle Chypotenuse=BCAC=k2k=12Now,(i) sinAcosC+cosAsinC=12×12+32×32=1(ii) cosAcosCsinAsinC=32×1212×32=0

Q.10

In Δ PQR, right-angled at Q, PR+QR=25 cm and PQ=5 cm. Determine the values of sin P, cos P and tan  P.

Ans.

Given that, PR+QR=25 cm and PQ=5 cmLet PR be x. Then, QR =25xApplying Pythagoras Theorem in Δ PQR, we get PR2=PQ2+QR2or x2=(5)2+(25x)2or x2=25+625+x250xor  50x=650or  x=13PR=x=13 cmand QR=25x=2513=12 cmsin P=side opposite to angle Phypotenuse=QRPR=1213cos P=side adjacent to angle Phypotenuse=PQPR=513tan P=side opposite to angle Pside adjacent to angle P=QRPQ=125

Q.11

State whether the following are true or false. Justifyyour answer.(i) The value of tan A is always less than 1.(ii) sec A=125 for some value of angle A.(iii) cos A is the abbreviation used for the cosecant of angle A.(iv) cot A is the product of cot and A.(v) sin θ=43 for some angle θ.

Ans.

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

tan A=side opposite to angle Aside adjacent to angle A=125>1tan A>1So, tan A<1 is not always true.Hence, the given statement is false.

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

We consider on the above Δ ABC, right-angled at B.sec A=hypotenuseside adjacent to angle A=ACAB=125Now, by using Pythagoras theorem, we have BC2=AC2AB2=14425=119or BC=119=10.9Therefore, sec A=125 is possible for some valuesof angle A.Hence, the given statement is true.(iii)The abbreviation used for the cosecant of angle A is cosec A.Therefore, the given statement is false.(iv) cot A is not the product of cot and A. Therefore, the givenstatement is true.(v) sinθ=side opposite to angle θhypotenuseIn a right-angled triangle hypotenuse is the longest side.Thus, sinθ=side opposite to angle θhypotenuse<1Therefore, sinθ=43>1 is not possible for any value of θ.Hence, the given statement is false.

Q.12

Evaluate the following:(i) sin 60° cos 30°+sin 30° cos 60°(ii) 2tan245°+cos230°sin260°(iii)cos 45°sec 30°+cosec 30°(iv) sin 30°+tan 45°cosec 60°sec 30°+cos 60°+cot 45°(v)  5cos260°+4sec230°tan245°sin230°+cos230°

Ans.

(i) sin 60° cos 30°+sin 30° cos 60°=32×32+12×12=1(ii)2tan245°+cos230°sin260°=2×(1)2+(32)2(32)2=2(iii)cos 45°sec 30°+cosec 30°=1223+2=322(1+3)                                           =322(1+3)×(13)(13)                                          =3322(13)=3342(iv) sin 30°+tan 45°cosec 60°sec 30°+cos 60°+cot 45°=12+12323+12+1                                                              =322323+32=33433+4 =33433+4×334334                                                            =(334)2(33)216=(33)2243+162716                                                            =4324311(v)  5cos260°+4sec230°tan245°sin230°+cos230°=5×(12)2+4×(23)21(12)2+(32)2                                                                =54+163114+34=15+64121244=6712

Q.13

Choose the correct option and justify your choice:(i) 2tan30°1+tan230°= (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°(ii) 1tan245°1+tan245°= (A) tan 90° (B) 1 (C) sin 45° (D) 0(iii) sin2A=2sinA is true when A= (A) 0° (B) 30° (C) 45° (D) 60°(iv) 2tan30°1tan230°= (A) cos 60° (B)sin 60° (C) tan 60° (D) sin 30

Ans.

(i)2tan30°1+tan230°=2×131+(13)2=233+13=2×343=32=sin60°Hence, the correct option is (A).(ii) 1tan245°1+tan245°=111+1=0Hence, the correct option is (D).(iii)sin2A=2sinAOn putting A=0°, we get sin 0°=2sin 0°or 0=2×0or 0=0Therefore, the correct option is (A).Other given options does not satisfy the given condition.(iv) 2tan30°1tan230°=2×131(13)2=23313=23×32=3=tan60°Therefore, the correct option is (C).

Q.14

If tan (A+B)=3 and tan (AB)=13; 0° < A + B90°; A > B, find A and B.

Ans.

We have       tan (A+B)=3or tan (A+B)=tan 60°or A+B=60° ...(1)Again, we have      tan (AB)=13or tan (AB)=tan 3or AB=30° ...(2)On adding equations (1) and (2), we get      2A=90°or A=45°On putting this value of A in equation (1), we get B=15°

Q.15

State whether the following are true or false. Justify your answer.(i) sin (A+B)=sin A+sin B.(ii) The value of sinθ increases as θ increases.(iii) The value of cosθ increases as θ increases.(iv) sinθ=cosθ for all values of θ.(v) cot A is not defined for A=0.

Ans.

(i)Let A=30° and B=60°.Now,sin(A+B)=sin(30°+60°)=sin90°=1andsin A+sin B=sin 30°+sin  60°=12+32=1+32sin(A+B)sin A+sin BHence, the given statement is false.(ii) We know thatsin0°=0,sin30°=12=0.5,sin45°=12=0.7,sin60°=32=0.87,sin90°=1We see that the value of sinθ increases as θ increasesin the interval 0°θ90°.Hence, the given statement is true if θ lies in theinterval 0°θ90°.(iii) We know thatcos0°=1,cos30°=32=0.87,cos45°=12=0.7,cos60°=12=0.5,cos90°=0We see that the value of cosθ decreases as θ increasesin the interval 0°θ90°.Hence, the given statement is false if θ lies in theinterval 0°θ90°.(iv) We know thatsin0°=0 and cos0°=1,sin30°=12=0.5 and cos30°=32=0.87.Therefore, sinθcosθ for all values of θ.Hence, the given statement is false.(v)We know that cot A=cosAsinA. So, for A=0, we havecot 0=cos0sin0=10We know that division by 0 is not defined. Therefore, cot 0is not defined.Hence, the given statement is true.

Q.16

Evaluate :(i) sin 18°cos 72° (ii) tan 26°cot 64° (iii) cos 48°sin 42° (iv) cosec 31°sec 59°

Ans.

(i) sin 18°cos 72°=cos(90°18°)cos 72°=cos 72°cos 72°=1(ii) tan 26°cot 64°=cot (90°26°)cot 64°=cot 64°cot 64°=1(iii) cos 48°sin 42°=sin (90°48°)sin 42°                                         =sin 42°sin 42°                                         =0(iv) cosec 31°sec 59°=sec (90°31°)sec 59°                                                =sec 59°sec 59°   =0

Q.17

Show that : (i) tan 48° tan 23° tan 42° tan 67°=1 (ii) cos 38° cos 52°sin 38° sin 52°=0

Ans.

(i) tan 48° tan 23° tan 42° tan 67°=cot (90°48°) cot (90°23°) tan 42° tan 67°=cot 42°  cot 67°  tan 42°  tan 67°=cot 42° tan 42°  tan 67° cot 67°=1×1=1(ii)cos 38° cos 52°sin 38° sin 52°=sin(90°38°) cos (90°38°)sin 38° sin 52°=sin  52° sin38°sin 38° sin 52°=0

Q.18

If tan 2A=cot(A18°), where 2A is an acute angle,find the value of A.

Ans.

Given that, tan 2A=cot(A18°)or cot (90°2A)=cot(A18°)or 90°2A=A18°or 3A=108°or A=108°3=36°

Q.19

If tan A=cot B, prove that A+B=90°.

Ans.

Given that,     tan A=cot Bor cot (90°A)=cot  Bor 90°A=Bor A+B=90°

Q.20

If sec 4A=cosec (A20°), where 4A is an acute angle, find the value of A.

Ans.

Given that, sec 4A=cosec (A20°)or cosec (90°4A)=cosec (A20°)or 90°4A=A20°or 5A=110°or A=110°5=22°

Q.21

If A, B and C are interior angles of a triangle ABC,then show thatsin(B+C2)=cosA2.

Ans.

Given that A, B and C are interior angles of a triangle ABC. A+B+C=180°or A=180°BCNow,sin(B+C2)=cos(90°B+C2)           =cos(180°BC2) =cos(A2)

Q.22

Express sin 67°+cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Ans.

sin 67°+cos 75°=cos(90°67°)+sin (90°75°) =cos 23°+sin 15°

Q.23

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Ans.

We know that cosec2 A=1+cot2 Aor    1sin2A=1+cot2 Aor   sin2A=11+cot2 Aor  sinA=11+cot2 AAlso, we know that sec2 A=1+tan2 Aor sec2 A=1+1cot2 Aor sec A=1+cot2Acot2A=1+cot2AcotAAlso, we know that tan A=sinAcosA=1cosAsinA=1cotA

Q.24

Write all the other trigonometric ratios of A in terms of sec A.

Ans.

We know that sin2 A=1cos2 Aor   sin2 A=11sec2 Aor   sin2A=sec2 A1sec2 Aor  sinA=sec2 A1sec Aor 1cosecA=sec2 A1sec Aor cosecA=sec Asec2 A1Also, we know that sec A cos A=1or cos A=1sec AAlso, we know that sec2Atan2 A=1or tan2 A=sec2A1or tan A=sec2A1Also, we know that tan A cot A=1or cot A=1tan A=1sec2A1

Q.25

Evaluate: (i) sin2 63°+sin2 27°cos2 17° +cos2 73° (ii) sin 25° cos 65°+cos 25° sin 65°

Ans.

(i) sin2 63°+sin2 27°cos2 17° +cos2 73°=sin2 63°+cos2(90°27°)sin2(90°17°)+cos2 73°    =sin2 63°+cos2 63°sin2 73°+cos2 73°                                               =11=1(ii) sin 25° cos 65°+cos 25° sin 65°=sin 25° sin(90°65°)+cos 25° cos(90° 65°)=sin 25° sin 25°+cos 25° cos 25°=sin2 25°+cos225°=1

Q.26

Choose the correct option. Justify your choice.(i) 9 sec2A9 tan2A= (A) 1 (B) 9 (C) 8 (D) 0(ii) (1 + tan θ+ sec θ) (1 + cotθcosecθ)= (A) 0 (B) 1 (C) 2 (D) 1(iii) (sec A + tan A) (1sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A(iv) 1+tan2A1+cot2 A= (A) sec2A (B) 1 (C) cot2A (D) tan2A

Ans.

(i) 9 sec2A9 tan2A=9(sec2Atan2A)=9×1=9Therefore, the correct option is (B).(ii) (1+tanθ+sec θ) (1+cotθcosecθ)=(1+sinθcosθ+1cosθ)(1+cosθsinθ1sinθ)=(sinθ+cosθ+1cosθ)(sinθ+cosθ1sinθ)=(sinθ+cosθ)21sinθcosθ=sin2θ+cos2θ+2sinθcosθ1sinθcosθ=1+2sinθcosθ1sinθcosθ=2Therefore, the correct option is (C).(iii) (sec A + tan A) (1sin A)=(1cos A+sinAcos A)(1sin A)=(1+sinAcos A)(1sin A)=1sin2Acos A=cos2Acos A=cosATherefore, the correct option is (D).(iv)    1+tan2A1+cot2A=sec2Acosec2A=sin2Acos2A=tan2ATherefore, the correct option is (D).

Q.27

Prove the following identities, where the angles involvedare acute angles for which the expressions are defined.(i) (cosecθcotθ)2=1cosθ1+cosθ(ii) cosA1+sinA+1+sinAcosA=2secA(iii) tanθ1cotθ+cotθ1tanθ=1+secθcosecθ [Hint :Write the expression in terms of sinθ and cosθ.](iv) 1+secAsecA=sin2A1cosA [Hint: Simplify LHS and RHS separately](v) cosAsinA+1cosA+sinA1=cosec A+cot A, using the identity cosec2A=1+cot2 A.(vi) 1+sinA1sinA=secA+tanA(vii) sinθ2sin3θ2cos3θcosθ=tanθ(viii) (sin A + cosec A)2+(cos A+sec A)2=7+tan2A+cot2A(ix) (cosec Asin A) (sec Acos A)=1tanA+cotA [Hint: Simplify LHS and RHS separately](x) 1+tan2A1+cot2A=(1tanA1cot A)2=tan2A

Ans.

(i) LHS=(cosecθcotθ)2         =cosec2θ+cot2θ2cosecθ cotθ         =1sin2θ+cos2θsin2θ2cosθsin2θ         =1+cos2θ2cosθsin2θ         =(1cosθ)(1cosθ)1cos2θ=(1cosθ)(1cosθ)(1+cosθ)(1cosθ)        =1cosθ1+cosθ=RHS(ii) LHS=cosA1+sinA+1+sinAcosA =cos2A+(1+sinA)2(1+sinA)cosA =cos2A+sin2A+2sinA+1(1+sinA)cosA        =1+1+2sinA(1+sinA)cosA =2(1+sinA)(1+sinA)cosA=2secA=RHS (iii) LHS=tanθ1cotθ+cotθ1tanθ                 =sinθcosθ1cosθsinθ+cosθsinθ1sinθcosθ         =sinθcosθsinθcosθsinθ+cosθsinθcosθsinθcosθ         =sin2θcosθ(sinθcosθ)+cos2θsinθ(cosθsinθ)        =sin3θcos3θsinθcosθ(sinθcosθ)      =(sin2θ+cos2θ+sinθcosθ)(sinθcosθ)sinθcosθ(sinθcosθ) =1+sinθcosθsinθcosθ=1+secθcosecθ=RHS(iv)LHS=1+secAsecA=1+1cosA1cosA=1+cosA1×1cosA1cosA =1cos2A1cosA=sin2A1cosA=RHS(v) LHS=cosAsinA+1cosA+sinA1         =cosAsinAsinAsinA+1sinAcosAsinA+sinAsinA1sinA         =cotA1+cosecAcotA+1cosec A         =cotA(1cosecA)cotA+(1cosec A)×cotA(1cosec A)cotA(1cosec A)         =cot2A+(1cosecA)22cotA(1cosec A)cot2A(1cosec A)2         =cot2A+1+cosec2A2cosecA2cotA+2cotAcosecAcot2A1cosec2 A+2cosecA    =2cosec2A2cosecA2cotA+2cotAcosecA11+2cosecA         =cosecA(cosecA1)+cotA(cosecA1)cosec A1    =cosecA+cot A(vi)LHS=1+sinA1sinA=1+sinA1sinA×1+sinA1+sinA =(1+sinA)11sin2A=1+sinAcosA=tanA+secA=RHS(vii) LHS=sinθ2sin3θ2cos3θcosθ         =sinθ(12sin2θ)cosθ(2cos2θ1)         =sinθ(12sin2θ)cosθ(2(1sin2θ)1)         =sinθ(12sin2θ)cosθ(22sin2θ1)         =sinθ(12sin2θ)cosθ(12sin2θ)         =tanθ=RHS(viii) LHS=(sin A+cosec A)2+(cos A+sec A)2   =sin2A+cosec2A+2+cos2A+sec2A+2 =sin2A+cos2A+4+sec2A+cosec2A =1+4+1+tan2A+1+cot2A         =7+tan2A+cot2A=RHS(ix) LHS=(cosec Asin A) (sec Acos A)         =(1sinAsinA)(1cosAcosA)         =(1sin2AsinA)(1cos2AcosA)         =cos2AsinA×sin2AcosA=sinAcosA1=sinAcosAsin2A+cos2A         =1sin2A+cos2AsinAcosA=1sinAcosA+cosAsinA        =1tanA+cotA        =RHS(x) LHS=1+tan2A1+cot2A=sec2Acosec2A=1cos2A1sin2A=sin2Acos2A=tan2A=RHSLHS=(1tanA1cot A)2=(1sinAcosA1cosAsinA)2=(cosAsinAcosAsinAcosAsinA)2         =sin2Acos2A=tan2A=RHS

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FAQs (Frequently Asked Questions)
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

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5. Explain the properties of angles.

The  different properties of angles are:

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