RS Aggarwal Chapter 8 Trigonometric Identities Exercise 8A Class 10 Maths

Chapter Name

RS Aggarwal Chapter 8 Trigonometric Identities

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 8B
  • Exercise 8C

Related Study

NCERT Solutions for Class 10 Maths

Exercise 8A Solutions

1. (i) (1 – cos2 θ) cosec2θ = 1 

(ii) (1 + cot2θ) sin2 θ = 1 

Solution

(i) LHS = (1 − cos2 θ) cosec2 θ

= sin2 θ cosec2 θ (∵ cos2 θ + sin2 θ = 1)

= 1/cosec2 θ × cosec2 θ

= 1 

Hence, LHS = RHS 

(ii) LHS = (1 + cot2 θ) sin2 θ

= cosec2 θ sin2 θ (∵ cosec2 θ − cot2 θ = 1)

= 1/sin2 θ × sin2 θ

= 1 

Hence, LHS = RHS


2. (i) (sec2 θ – 1) cot2 θ = 1

(ii) (sec2 θ – 1)(cosec2 θ - 1) = 1

(iii) (1 – cos2 θ) sec2 θ = tan2 θ

Solution

(i) LHS = (sec2 θ − 1) cot2 θ 

= tan2 θ × cot2 θ (∵ sec2 θ − tan2 θ = 1) 

 = 1/cot2 θ × cot2 θ 

= 1 

= RHS 

(ii) LHS = (sec2 θ − 1)(cosec2 θ − 1) 

= tan2 θ × cot2 θ (∵ sec2 θ − tan2 θ = 1)

= 1/cot2 θ × cot2 θ
= 1 

= RHS


(iii) LHS = (1 - cos2 θ)sec2 θ

= sin2 θ × sec2 θ (∵ sin2 θ + cos2 θ = 1)

= sin2 θ × 1/cos2 θ

= sin2 θ/cos2 θ

= tan2 θ

= RHS


3. (i) sin2 θ + 1/(1 + tan2 θ) = 1

(ii) 1/(1 + tan2 θ) + 1/(1 + cot2 θ) = 1

Solution

(i) LHS = sin2 θ + 1/(1 + tan2 θ)

= sin2 θ + 1/sec2 θ  (∵ sec2 θ − tan2 θ = 1) 

= sin2 θ + cos2 θ

= 1 

= RHS 

(ii) LHS = 1/(1 + tan2 θ) + 1/(1 + cot2 θ) 

= 1/sec2 θ + 1/cosec2 θ

= cos2 θ + sin2 θ

= 1 

= RHS


4. (i) (1 + cos θ)(1 – cos θ)(1 + cos2 θ) = 1

(ii) cosec θ (1 + cos θ)(cosec θ – cot θ) = 1

Solution

(i) LHS = (1+ cos θ)/(1 – cos θ) (1 + cos2 θ)

= (1 – cos2 θ)(cosec2 θ)

= sin2 θ × cosec2 θ

= sin2 θ × 1/sin2 θ

= 1

= RHS

(ii) LHS = cosec θ(1 + cos θ)(cosec θ – cot θ)

= (cosec θ + cosec θ × cos θ)(cosec θ – cot θ)

= (cosec θ + 1/sin θ × cos θ)(cosec θ – cot θ)

= (cosec θ + cot θ)(cosec θ – cot θ)

= cosec2 θ – cot2 θ (∵ cosec2 θ - cot2 θ = 1)

= 1

= RHS


5. (i) cot2 θ - 1/sin2 θ = -1

(ii) tan2 θ – 1/cos2 θ = - 1

(iii) cos2 θ + 1/(1 + cot2 θ) = 1

Solution

(i) cot2 θ - 1/sin2 θ = -1

= cos2 θ/sin2 θ – 1/sin2 θ

= (cos2 θ - 1)/sin2 θ

= - sin2 θ /sin2 θ

= - 1

= RHS

(ii) LHS = tan2 θ – 1/cos2 θ

= sin2 θ/cos2 θ – 1/cos2 θ

= (sin2 θ – 1)/cos2 θ

= -cos2 θ/cos2 θ 

= - 1

= RHS

(iii) LHS = cos2 θ + 1/(1 + cot2 θ)

= cos2 θ + 1/cosec2 θ

= cos2 θ + sin2 θ

= 1

= RHS


6. 1/(1 + sin θ) + 1/(1 – sin θ) = 2 sec2 θ

Solution

LHS = 1/(1 + sin θ) + 1/(1 – sin θ)

= (1 – sin θ) + (1 + sin θ) /(1 + sin θ)(1 – sin θ)

= 2/(1 – sin2 θ)

= 2/cos2 θ

= 2 sec2 θ

= RHS


7. (i) sec θ(1 – sin θ)(sec θ + tan θ) = 1

(ii) sin θ (1 + tan θ) + cos θ(1 + cot θ) = sec θ + cosec θ)

Solution

(i) LHS = sec θ(1 – sin θ)(sec θ + tan θ)

= (sec θ – sec θ sin θ)(sec θ + tan θ)

= (sec θ – 1/cos θ × sin θ) (sec θ + tan θ)

= (sec θ – tan θ)(sec θ + tan θ)

= sec2 θ – tan2 θ

= 1

= RHS

(ii) LHS = sin θ(1 + tan θ) + cos θ(1 + cot θ)

= sin θ + sin θ × sin θ/cos θ + cos θ × cos θ/sin θ

= (cos θ sin2 θ + sin3 θ + cos2 θ sin θ + cos3 θ)/(cos θ sin θ)

= (sin3 θ + cos3 θ) + (cos θ sin2 θ + cos2 θ sin θ)/(cos θ sin θ)

= [(sin θ + cos θ)(sin2 θ – sin θ cos θ + cos2 θ) + sin θ cos θ (sin θ + cos θ)]/cos θ sin θ

= (sin θ + cos θ)(sin2 θ + cos2 θ – sin θ cos θ + sin θ cos θ)/(cos θ sin θ)

= (sin θ + cos θ)(1)/(cos θ sin θ)

= (sin θ)/(cos θ sin θ) + (cos θ)/(cos θ sin θ)

= 1/cos θ + 1/sin θ

= sec θ + cosec θ

= RHS


8. (i) 1 + cot2 θ/(1 + cosec θ) = cosec θ

(ii) 1 + tan2 θ/(1 + sec θ) = sec θ

Solution

(i) LHS = 1 + cot2 θ/(1 + cosec θ)

= 1 + (cosec2 θ – 1)/(cosec θ + 1) (∵ cosec2 θ – cot2 θ = 1)

= 1 + (cosec θ + 1)(cosec θ – 1)/(cosec θ + 1)

= 1 + (cosec θ – 1)

= cosec θ

= RHS

(ii) LHS = 1 + tan2 θ/(1 + sec θ)

= 1 + (sec2 θ – 1)/(sec θ + 1)

= 1 + {(sec θ + 1)(sec θ – 1)}/(sec θ + 1)

= 1 + (sec θ – 1)

= sec θ

= RHS


9. 1 + (tan2 θ) cot θ/(cosec2 θ) = tan θ

Solution

LHS = (1 + tan2 θ)cot θ/cosec2 θ

= (sec2 θ cot θ)/(cosec2 θ)

= (1/cos2 θ × cos θ/sin θ)/(1/sin2 θ)

= 1/(cos θ sin θ) × sin2 θ

= sin θ/cos θ

= tan θ

= RHS

Hence, LHS = RHS


10. tan2 θ/(1 + tan2 θ) + cot2 θ/(1 + cot2 θ) = 1

Solution

LHS = tan2 θ/(1 + tan2 θ) + cot2 θ/(1 + cot2 θ)

= tan2 θ/sec2 θ + cot2 θ/cosec2 θ (∵ sec2 θ – tan2 θ = 1 and cosec2 θ – cot2 θ = 1)

= (sin2 θ/cos2 θ)/(1/cos2 θ) + (cos2 θ/sin2 θ)/(1/sin2 θ)

= sin2 θ + cos2 θ

= 1

= RHS

Hence, LHS = RHS


11. sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2 cosec θ

Solution

LHS = sin (1 + cos θ) + (1 + cos θ)/sin θ

= sin2 θ + (1 + cos θ)2/(1 + cos θ) sin θ

= sin2 θ + 1 + cos2 θ + 2 cos θ)/)(1 + cos θ) sin θ

= (1 + 1 + 2 cos θ)/(1 + cos θ) sin θ

= (2 + 2 cos θ)/(1 + cos θ) sin θ

= 2(1 + cos θ)/(1 + cos θ) sin θ

= 2/sin θ

= 2 cosec θ

= RHS

Hence, L.H.S = R.H.S


12. tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = (1 + sec θ cosec θ)

Solution

tan θ/(1 – cot θ) + cot θ/(1 – tan θ)


= sec θ cosec θ + 1

= 1 + sec θ cosec θ

= RHS


13. cos2 θ/(1 – tan θ) + sin3 θ/(sin θ – cos θ) = (1 + sin θ cos θ)

Solution

cos2 θ/(1 – tan θ) + sin3 θ/(sin θ – cos θ) = (1 + sin θ cos θ)

LHS = cos2 θ/(1 – tan θ) + sin3 θ/(sin θ – cos θ)

= (sin2 θ + cos2 θ + cos θ sin θ)

= (1 + sin θ cos θ)

= RHS

Hence, L.H.S. = R.H.S.


14. cos θ/(1 – tan θ) + sin2 θ/(cos θ – sin θ)

Solution

LHS = cos θ/(1 – tan θ) – sin2 θ/(cos θ – sin θ)

= (cos θ + sin θ)

= RHS

Hence, LHS = RHS


15. (1 + tan2 θ)(1 + cot2 θ) = 1/(sin2 θ – sin4 θ)

Solution

L.H.S. = (1 + tan2 θ)(1 + cot2 θ)

= sec2 θ. cosec2 θ  (∵ sec2 θ – tan2 θ = 1 and cosec2 θ – cot2 θ = 1)

= 1/cos2 θ. sin2 θ

= 1/(1 – sin2 θ)sin2 θ

= 1/sin2 θ – sin4 θ

= RHS

Hence, LHS = RHS


16. tan θ/(1 + tan2 θ)2 + cot θ/(1 + cot2 θ)2 = sin θ cos θ

Solution

L.H.S. = tan θ/(1 + tan2 θ) + cot θ/(1 + cot2 θ)2

= tan θ/(sec2 θ)2 + cot θ/(cosec2 θ)2

= tan θ/sec4 θ + cot θ/cosec4 θ

= sin θ/cos θ × cos4 θ + cos θ/sin θ × sin4 θ

= sin θ cos3 θ + cos θ sin3 θ

= sin θ cos θ (cos2 θ + sin2 θ)

= sin θ cos θ

= RHS


17. (i) sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2 θ

(ii) sin2 θ + cos4 θ = cos2 θ + sin4 θ

(iii) cosec4 θ + cosec2 θ = cot4 θ + cot2 θ

Solution

(i) LHS = sin6 θ + cos6 θ

= (sin2 θ)3 + cos2 θ)3

= (sin2θ + cos2 θ)(sin4 θ – sin2 θ cos2 θ + cos4 θ)

= 1 × {(sin2 θ)2 + 2 sin2 θ cos2 θ + (cos2 θ)2 – 3 sin2 θ cos2 θ}

= (sin2 θ + cos2 θ)2 – 3 sin2 θ cos2 θ

= (1)2 – 3 sin2 θ cos2 θ

= 1 – 3 sin2 θ cos2 θ

= RHS

Hence, LHS = RHS

(ii) LHS = sin2 θ + cos4 θ

= sin2 θ + (cos2 θ)2

= sin2 θ + (1 – sin2 θ)2

= sin2 θ + 1 – 2 sin2 θ + sin4 θ

= 1 – sin2 θ + sin4 θ

= cos2 θ + sin4 θ

= RHS

Hence, LHS = RHS

(iii) LHS = cosec4 θ – cosec2 θ

= cosec2 θ(cosec2 θ – 1)

= cosec2 θ × cot2 θ (∵ cosec2 θ – cot2 θ = 1)

= (1 + cot2 θ) × cot2 θ

= cot2 θ + cot4 θ

= RHS

Hence, LHS = RHS


18. (i) (1 – tan2 θ)/(1 + tan2 θ) = (cos2 θ – sin2 θ)

(ii) (1 – tan2 θ)/(cot2 – 1) = tan2 θ

Solution

(i) LHS = (1 – tan2 θ)/(1 + tan2 θ)

= (cos2 θ – sin2 θ)/(cos2 θ + sin2 θ)

= (cos2 θ – sin2 θ)/1

= cos2 θ – sin2 θ

= RHS

(ii) LHS = (1 – tan2 θ)/(cot2 θ – 1)

= tan2 θ

= RHS


19. (i) tan θ/(sec θ – 1) + tan θ/(sec θ + 1) = 2 cosec θ

(ii) cot θ/(cosec θ + 1) + cosec θ + 1)/cot θ) = 2 sec θ

Solution

(i) LHS = tan θ/(sec θ – 1) + tan θ/(sec θ + 1)

= RHS
Hence, LHS = RHS 

(ii) LHS = cot θ/(cosec θ + 1) + (cosec θ + 1)/cot θ

= 2 × 1/sin θ × sin θ/cos θ

= 2 sec θ

= RHS

Hence, LHS = RHS


20. (i) (sec θ – 1)/(sec θ + 1) = sin2 θ/(1 + cos θ)2

(ii) (sec θ – tan θ)/(sec θ + tan θ) = cos2 θ/(1 + sin θ)2

Solution

(i) LHS = (sec θ – 1)/(sec θ + 1)


(ii) LHS = (sec θ – tan θ)/(sec θ + tan θ)

= (1 - sin2 θ)/(1 + sin θ)2

= cos2 θ/(1 + sin θ)2

= RHS


21. (i) √(1+sinθ)/(1-sinθ) = (secθ + tanθ)

(ii) √(1-cosθ)/(1+cosθ) = (cosecθ - cotθ)

(iii) √(1+cosθ)/(1-cosθ) = 2cosecθ 

Solution

(i) LHS 

= (1 + sin θ)/cos θ

= 1/cos θ + sin θ/cos θ

= (sec θ + tan θ)

= RHS

(ii) LHS 

(iii) 

= (1 + cos θ + 1 - cos θ)/sin θ

= 2/sin θ

= 2 cosec θ

= RHS


22. (cos3 θ + sin3 θ)/(cos θ + sin θ) + (cos3 θ - sin3 θ)/(cos θ – sin θ) = 2

Solution

LHS = (cos3 θ + sin3 θ)/(cos θ + sin θ) + (cos3 θ – sin3 θ)/(cos θ – sin θ)

= (cos2 θ + sin2 θ – cos θ sin θ) + (cos2 θ + sin2 θ + cos θ sin θ)

= (1 – cos θ sin θ) + (1 + cos θ sin θ)

= 2

= RHS

Hence, LHS = RHS


23. sin θ/(cot θ + cosec θ) – sin θ/(cot θ – cosec θ) = 2

Solution

LHS = sin θ/(cot θ + cosec θ) – sin θ/(cot θ – cosec θ)

= sin θ × 2 × 1/sin θ

= 2

= RHS


24. (i) (sin θ – cos θ)/(sin θ + cos θ) + (sin θ + cos θ)/(sin θ – cos θ) = 2/(2 sin2 θ – 1)

(ii) (sin θ + cos θ)/(sin θ – cos θ) + (sin θ – cos θ)/(sin θ + cos θ) = 2/(1 – 2 cos2 θ)

Solution

(i) LHS = (sin θ – cos θ)/(sin θ + cos θ) + (sin θ + cos θ)/(sin θ – cos θ)

(ii) LHS = (sin θ + cos θ)/(sin θ - cos θ) + (sin θ - cos θ)/(sin θ + cos θ)


25. (1 + cos θ - sin2θ)/sin θ(1 + cos θ) = cot θ

Solution

LHS = (1 + cos θ – sin2 θ)/{sin θ(1 + cos θ)}

= (1 + cos θ) – (1 – cos2 θ)/sin θ(1 + cos θ)

= (cos θ + cos2 θ)/sin θ(1 + cos θ)

= cos θ(1 + cos θ)/sin θ(1 + cos θ)

= cos θ/sin θ

= cot θ

= RHS

Hence, L.H.S = R.H.S.


26. (i) (cosec θ + cot θ)/(cosec θ – cot θ) = (cosec θ + cot θ)2 = 1 + 2 cot2 θ + 2 cosec θ cot θ

(ii) (sec θ + tan θ)/(sec θ – tan θ)2 = (sec θ + tan θ)2 = 1 + 2 tan2 θ + 2 sec θ tan θ

Solution

(i) Here, (cosec θ + cot θ)/(cosec θ – cot θ)

= (cosec θ + cot θ)(cosec θ + cot θ)/(cosec θ – cot θ)(cosec θ + cot θ)

= (cosec θ + cot θ)2/(cosec2 θ – cot2)

= (cosec θ + cot θ)2/1

= (cosec θ + cot θ)2/1

= (cosec θ + cot θ)2

Again, (cosec θ + cot θ)2

= cosec2 θ + cot2 θ + 2cosec θ cot θ

= 1 + cot2 θ + cot2 θ + 2 cosec θ cot θ (∵ cosec2 θ – cot2 θ = 1)

= 1 + 2 cot2 θ + 2 cosec θ cot θ

(ii) Here, (sec θ + tan θ)/(sec θ – tan θ)

= (sec θ + tan θ)(sec θ + tan θ)/(sec θ + tan θ)(sec θ + tan θ)

= (sec θ + tan θ)2/(sec θ – tan2 θ)

= (sec θ + tan θ)2/1

= (sec θ + tan θ)2

Again, (sec θ + tan θ)2

= sec2 θ + tan2 θ + 2 sec θ tan θ

= 1 + tan2 θ + tan2 θ + 2sec θ tan θ

= 1 + 2 tan2 θ + 2 sec θ tan θ


27. (i) (1 + cos θ + sin θ)/(1 + cos θ – sin θ) = (1 + sin θ)/cos θ

(ii) (sin θ + 1 – cos θ)/(cos θ – 1 + sin θ) = (1 + sin θ)/cos θ

Solution

(i) LHS = (1 + cos θ + sin θ)/(1 + cos θ – sin θ)


(ii) LHS = (sin θ + 1 cos θ)/(cos θ – 1 + sin θ)

= (2 sin2 θ + 2 sin θ)/(2 sin θ cos θ)

= 2 sin θ(1 + sin θ)/2 sin θ cos θ

= (1 + sin θ)/cos θ

= RHS


28. sin θ/(sec θ + tan θ – 1) + cos θ/(cosec θ + cot θ – 1) = 1

Solution

LHS = sin θ/(sec θ + tan θ + 1) + cos θ/(cosec θ + cot θ – 1)

= (sin θ cos θ)/(1 + sin θ – cos θ) + cos θ sin θ/(1 + cos θ – sin θ)


= RHS 

Hence, LHS = RHS


29. (sin θ + cos θ)/(sin θ – cos θ) + (sin θ – cos θ)/(sin θ + cos θ) = 2/(sin2 θ – cos2 θ) = 2/(sin2 θ – 1)

Solution

We have (sin θ + cos θ)/(sin θ – cos θ) + (sin θ – cos θ)/(sin θ + cos θ)

= (sin θ + cos θ)2 + (sin θ – cos θ)2/(sin θ – cos θ)(sin θ + cos θ)

= (sin2 θ + cos2 θ + 2 sin θ cos θ + sin2 θ + cos2 θ – 2 sin θ cos θ)/(sin2 θ – cos2 θ)

= (1 + 1)/(sin2 θ – cos2 θ)

= 2/(sin2 θ – cos2 θ)

Again, 2/(sin2 θ – cos2 θ)

= 2/(sin2 θ – (1 – sin2 θ))

= 2/(2 sin2 θ – 1)


30. (cos θ cosec θ – sin θ sec θ)/(cos θ + sin θ) = cosec θ – sec θ

Solution

LHS = (cos θ cosec θ – sin θ sec θ)/(cos θ + sin θ)

= 1/sin θ – 1/cos θ

= cosec θ – sec θ

= RHS

Hence, LHS = RHS 


31. (1 + tan θ + cot θ)(sin θ – cos θ) = (sec θ/cosec2 θ - cosec θ/sec2 θ)

Solution

LHS = (1 + tan θ + cot θ)(sin θ – cos θ)

= sin θ + tan θ sin θ + cot θ sin θ – cos θ – tan θ cos θ – cot θ cos θ

= sin θ + tan θ sin θ + cos θ/sin θ × sin θ – cos θ – sin θ/cos θ × cos θ – cot θ cos θ

= sin θ + tan θ sin θ + cos θ – cos θ – sin θ – cot θ cos θ

= tan θ sin θ – cot θ cos θ

= sin θ/cos θ × 1/cosec θ × sec θ – 1/sec θ × 1/sec θ × cosec θ

= sec θ/cosec2 θ – cosec θ/sec2 θ

= RHS

Hence, LHS = RHS


32. cot2 θ(sec θ – 1)/(1 + sin θ) + sec2 θ(sin θ – 1)/(1 + sec θ) = 0

Solution

LHS = cot2 θ(sec θ – 1)/(1 + sin θ) + sec2 θ(sin θ – 1)/(1 + sec θ)

= θ
= RHS


33. {1/(sec2 θ – cos2 θ) + 1/(cosec2 θ – sin2 θ)} (sin2 θ cos2 θ) = (1 – sin2 θ cos2 θ/(2 + sin2 θ cos2 θ)

Solution

LHS = {1/sec2 θ – cos2 θ + 1/cosec2 θ – sin2 θ)}(sin2 θ cos2 θ)


34. (sin A – sin B)/(cos A + cos B) + (cos A – cos B)/(sin A + sin B) = 0

Solution

LHS = (sin A – sin B)/(cos A + cos B) + (cos A – cos B)/(sin A + sin B)

= (sin A – sin B)(sin A + sin B) + (cos A – cos B)(cos A – cos B) /(cos A + cos B)(sin A + sin B)

= sin2 A – sin2 B + cos2 A – cos2 B)/(cos A + cos B)(sin A + sin B)

= 0/(cos A + cos B)(sin A + sin B)

= 0

= RHS


35. (tan A + tan B)/(cot A + cot B) = tan A tan B

Solution

LHS = (tan A + tan B)/(cot A + cot B)

= tan A tan B

= RHS

Hence, LHS = RHS


36. Show that none of the following is an identity:

(i) cos2 θ + cos θ = 1

(ii) sin2 θ + sin θ = 2

(iii) tan2 θ + sin θ = cos2 θ 

Solution

(i) cos2 θ + cos θ = 1

LHS = cos2 θ + cos θ 

= 1 – sin2 θ + cos θ 

= 1 – (sin2 θ - cos θ)

Since, LHS ≠ RHS, this not an identity.

(ii) sin2 θ + sin θ = 1

LHS = sin2 θ + sin θ 

= 1 – cos2 θ + sin θ 

= 1 – (cos2 θ - sin θ)

Since LHS ≠ RHS, this not an identity.

(iii) tan2 θ + sin θ = cos2 θ 

LHS = tan2θ + sin θ 

= sin2 θ /cos2 θ + sin θ 

= (1 – cos2 θ)/cos2 θ + sin θ 

= sec2 θ - 1 + sin θ 

Since LHS ≠ RHS, this is not an identity.


37. Prove that (sin θ - 2 sin3 θ) = (2 cos3 θ - cos θ) tan θ 

Solution

RHS = (2 cos3 θ – cos θ) tan θ

= (2 cos2 θ – 1) cos θ × sin θ/cos θ

= [2(1 – sin2 θ) – 1] sin θ

= (2 – 2 sin2 θ – 1) sin θ

= (1 – 2 sin2 θ) sin θ

= (sin θ – 2 sin3 θ)

= LHS

Previous Post Next Post