NCERT Solutions Class 10 Maths Chapter 8

NCERT Solutions Class 10 Mathematics Chapter 8: Introduction to Trigonometry 

Chapter 8 in the Class 10 CBSE syllabus introduces students to the concept of Trigonometry. Students can subscribe to  NCERT Solutions Class 10 Mathematics Chapter 8 by Extramarks so that they can study from it independently without any assistance from teachers or parents. It will ensure that even the minutest doubts are resolved. In fact, the students would be interested in learning and mastering the topic with ease.. The subject matter experts have written the answers in the solutions while ensuring that they are accurate and adhere to  the guidelines laid down by CBSE.

NCERT Solutions Class 10 Mathematics Chapter 8: Introduction to Trigonometry 

Access NCERT Solutions for Class 10 Mathematics Chapter 8 – Introduction to Trigonometry

NCERT Solutions for Class 10 Mathematics Chapter 8 –

Trigonometry can be a very challenging topic to master for students since it involves so many different formulas and theorems that students need to know and apply them correctly. Furthermore, there are many different kinds of problems that can be asked from this topic. NCERT Solutions for Class 10 Mathematics Chapter 8 will, therefore, be an extremely useful resource as students will find step-by-step explanations to all the problems given in the NCERT textbook. 

NCERT Solutions for Class 10 Mathematics

Students can also access detailed solutions to all the chapters covered in the Class 10 Mathematics textbook listed below:

Chapter 1: Real Numbers 

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables 

Chapter 4: Quadratic Equations 

Chapter 5: Arithmetic Progressions 

Chapter 6: Triangles 

Chapter 7: Coordinate Geometry 

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry 

Chapter 10: Circles 

Chapter 11: Constructions 

Chapter 12: Areas Related to Circles 

Chapter 13: Surface Areas and Volumes 

Chapter 14: Statistics 

Chapter 15: Probability 

About the Chapter – Introduction to Trigonometry

Trigonometry is a Mathematics concept that focuses on triangles. Trigonometry is a blend of ‘trigono’ and ‘metry’, which means triangle and measure. 

The chapter includes proper methods and formulae to help the students find the missing sides and angle of a triangle.  Class 10 chapter 8 has been segmented into five sections. 

The first section presents the problems depicted using the right-angle triangle and the second section comprises an introduction to trigonometric ratios with examples. The students can work on the exercises to understand the topics in detail. The segment also explains the derivation of sine, cosine, and several other trigonometric functions.

The third section talks about the trigonometric ratios of measurement. The fourth section presents a few solved examples, trigonometric ratio criteria for the complementary angles and the exercises. The last section explains the subject associated with the trigonometric identities, accompanied by examples and exercises. 

Benefits of NCERT Solutions for Class 10 Mathematics

NCERT Solutions for Class 10 Mathematics can be extremely beneficial for the students to master the topic and increase their confidence in achieving a higher grade.  

   The multiple advantages of studying NCERT solutions are:

  • The solutions have been created by experienced faculty and subject-matter experts with years of experience thus students can be assured that they are referring to reliable and authentic learning material. 
  • The NCERT Solutions strictly follow the CBSE Guidelines.
  • Important Terms to Remember in Height and Distance
  • The solutions break down the complex trigonometric problems in a step by step manner to ease the students’ learning process.
  • The solutions are descriptive and to the point. .
  • The answers have been described, accompanied by Illustrations and examples.
  • The language used in the solutions is easy-to-understand.

Related Questions

Question: State whether the following condition is true or false: The value of sinθ increases as θ increases between 0 and 90 degrees.

Solution: We know the values of sine for different values of θ between 0 and 90 degrees

sin 0 = 0

sin 30 = 1/2 = 0.5

sin 45 = 1/√2 = 0.7071

sin 60 = √3/2 = 0.866

sin 90 = 1

It can be seen that the value of sinθ increases as θ increases between 0 and 90 degrees. Therefore, the given statement is true.

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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The Exercise 8.4 has questions related to trigonometric ratios of complementary angles. Students will have to memorise some standard formulae in order to solve every question from the exercise.