Understanding different trigonometric ratios and its relations between them to prove various trigonometric identities is the major focus in this exercise. Students who want to learn the right procedures to prove such identities can refer to the Selina Solutions for Class 10 Maths. The solutions are created by subject matter experts at BYJU’S. The solution PDF of the Concise Selina Solutions for Class 10 Maths Chapter 21 Trigonometrical Identities Exercise 21(A) is available in the link provided below.
Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(A) Download PDF
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Access Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(A)
Prove the following identities:
1. sec A – 1/ sec A + 1 = 1 – cos A/ 1 + cos A
Solution:
– Hence Proved
2. 1 + sin A/ 1 – sin A = cosec A + 1/ cosec A – 1
Solution:
– Hence Proved
3. 1/ tan A + cot A = cos A sin A
Solution:
Taking L.H.S,
– Hence Proved
4. tan A – cot A = 1 – 2 cos2 A/ sin A cos A
Solution:
Taking LHS,
– Hence Proved
5. sin4 A – cos4 A = 2 sin2 A – 1
Solution:
Taking L.H.S,
sin4 A – cos4 A
= (sin2 A)2 – (cos2 A)2
= (sin2 A + cos2 A) (sin2 A – cos2 A)
= sin2A – cos2A
= sin2A – (1 – sin2A) [Since, cos2 A = 1 – sin2 A]
= 2sin2 A – 1
– Hence Proved
6. (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A
Solution:
Taking L.H.S,
(1 – tan A)2 + (1 + tan A)2
= (1 + tan2 A + 2 tan A) + (1 + tan2 A – 2 tan A)
= 2 (1 + tan2 A)
= 2 sec2 A [Since, 1 + tan2 A = sec2 A]
– Hence Proved
7. cosec4 A – cosec2 A = cot4 A + cot2 A
Solution:
cosec4 A – cosec2 A
= cosec2 A(cosec2 A – 1)
= (1 + cot2 A) (1 + cot2 A – 1)
= (1 + cot2 A) cot2 A
= cot4 A + cot2 A = R.H.S
– Hence Proved
8. sec A (1 – sin A) (sec A + tan A) = 1
Solution:
Taking L.H.S,
sec A (1 – sin A) (sec A + tan A)
– Hence Proved
9. cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:
Taking L.H.S,
– Hence Proved
10. sec2 A + cosec2 A = sec2 A . cosec2 A
Solution:
Taking L.H.S,
– Hence Proved
11. (1 + tan2 A) cot A/ cosec2 A = tan A
Solution:
Taking L.H.S,
= RHS
– Hence Proved
12. tan2 A – sin2 A = tan2 A. sin2 A
Solution:
Taking L.H.S,
tan2 A – sin2 A
– Hence Proved
13. cot2 A – cos2 A = cos2A. cot2A
Solution:
Taking L.H.S,
cot2 A – cos2 A
– Hence Proved
14. (cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Solution:
Taking L.H.S,
(cosec A + sin A) (cosec A – sin A)
= cosec2 A – sin2 A
= (1 + cot2 A) – (1 – cos2 A)
= cot2 A + cos2 A = R.H.S
– Hence Proved
15. (sec A – cos A)(sec A + cos A) = sin2 A + tan2 A
Solution:
Taking L.H.S,
(sec A – cos A)(sec A + cos A)
= (sec2 A – cos2 A)
= (1 + tan2 A) – (1 – sin2 A)
= sin2 A + tan2 A = RHS
– Hence Proved
16. (cos A + sin A)2 + (cosA – sin A)2 = 2
Solution:
Taking L.H.S,
(cos A + sin A)2 + (cosA – sin A)2
= cos2 A + sin2 A + 2cos A sin A + cos2 A – 2cosA.sinA
= 2 (cos2 A + sin2 A) = 2 = R.H.S
– Hence Proved
17. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Solution:
Taking LHS,
(cosec A – sin A)(sec A – cos A)(tan A + cot A)
= RHS
– Hence Proved
18. 1/ sec A + tan A = sec A – tan A
Solution:
Taking LHS,
= RHS
– Hence Proved
19. cosec A + cot A = 1/ cosec A – cot A
Solution:
Taking LHS,
cosec A + cot A
= RHS
– Hence Proved
20. sec A – tan A/ sec A + tan A = 1 – 2 secA tanA + 2 tan2 A
Solution:
Taking LHS,
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 A = RHS
– Hence Proved
21. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Solution:
Taking LHS,
(sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2cos A sec A
= (sin2 A + cos2 A ) + cosec2 A + sec2 A + 2 + 2
= 1 + cosec2 A + sec2 A + 4
= 5 + (1 + cot2 A) + (1 + tan2 A)
= 7 + tan2 A + cot2 A = RHS
– Hence Proved
22. sec2 A. cosec2 A = tan2 A + cot2 A + 2
Solution:
Taking,
RHS = tan2 A + cot2 A + 2 = tan2 A + cot2 A + 2 tan A. cot A
= (tan A + cot A)2 = (sin A/cos A + cos A/ sin A)2
= (sin2 A + cos2 A/ sin A.cos A)2 = 1/ cos2 A. sin2 A
= sec2 A. cosec2 A = LHS
– Hence Proved
23. 1/ 1 + cos A + 1/ 1 – cos A = 2 cosec2 A
Solution:
Taking LHS,
= RHS
– Hence Proved
24. 1/ 1 – sin A + 1/ 1 + sin A = 2 sec2 A
Solution:
Taking LHS,
= RHS
– Hence Proved
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