RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities are provided in this article. The 6th Chapter of RD Sharma Class 10 is Trigonometric Identities, which has two exercises and is solved with thorough explanations here RD Sharma Solutions for Class 10. The preceding chapter discussed trigonometric ratios and their relationships. This chapter, on the other hand, will focus on establishing some trigonometric identities and applying them to the derivation of other relevant trigonometric identities.
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RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Page No: 6.43
Prove the following trigonometric identities:
1. (1 – cos2 A) cosec2 A = 1
Solution:
Taking the L.H.S.,
(1 – cos2 A) cosec2 A
= (sin2 A) cosec2 A [∵ sin2 A + cos2 A = 1 ⇒1 – sin2 A = cos2 A]
= 12
= 1 = R.H.S.
– Hence, proved.
2. (1 + cot2 A) sin2 A = 1
Solution:
By using the identity,
cosec2 A – cot2 A = 1 ⇒ cosec2 A = cot2 A + 1
Taking,
L.H.S. = (1 + cot2 A) sin2 A
= cosec2 A sin2 A
= (cosec A sin A)2
= ((1/sin A) × sin A)2
= (1)2
= 1
= R.H.S.
– Hence, proved.
3. tan2 θ cos2 θ = 1 − cos2 θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
Taking,
L.H.S. = tan2 θ cos2 θ
= (tan θ × cos θ)2
= (sin θ)2
= sin2 θ
= 1 – cos2 θ
= R.H.S.
– Hence, proved.
4. cosec θ √(1 – cos2 θ) = 1
Solution:
Using identity,
sin2 θ + cos2 θ = 1 ⇒ sin2 θ = 1 – cos2 θ
Taking L.H.S.,
L.H.S = cosec θ √(1 – cos2 θ)
= cosec θ √( sin2 θ)
= cosec θ x sin θ
= 1
= R.H.S.
– Hence, proved.
5. (sec2 θ − 1)(cosec2 θ − 1) = 1
Solution:
Using identities,
(sec2 θ − tan2 θ) = 1 and (cosec2 θ − cot2 θ) = 1
We have,
L.H.S. = (sec2 θ – 1)(cosec2θ – 1)
= tan2θ × cot2θ
= (tan θ × cot θ)2
= (tan θ × 1/tan θ)2
= 12
= 1
= R.H.S.
– Hence, proved.
6. tan θ + 1/ tan θ = sec θ cosec θ
Solution:
We have,
L.H.S. = tan θ + 1/ tan θ
= (tan2 θ + 1)/ tan θ
= sec2 θ / tan θ [∵ sec2 θ − tan2 θ = 1]
= (1/cos2 θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]
= cos θ/ (sin θ x cos2 θ)
= 1/ cos θ x 1/ sin θ
= sec θ x cosec θ
= sec θ cosec θ
= R.H.S.
– Hence, proved.
7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1+ sin θ), we get
L.H.S. = R.H.S.
– Hence, proved.
8. cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1- sin θ), we get
L.H.S. = R.H.S.
– Hence, proved.
9. cos2 θ + 1/(1 + cot2 θ) = 1
Solution:
We already know that,
cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1
Taking L.H.S.,
= cos2 A + sin2 A
= 1
= R.H.S.
– Hence, proved.
10. sin2 A + 1/(1 + tan 2 A) = 1
Solution:
We already know that,
sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1
Taking L.H.S.,
= sin2 A + cos2 A
= 1
= R.H.S.
– Hence, proved.
11.
Solution:
We know that,
sin2 θ + cos2 θ = 1
Taking the L.H.S.,
= cosec θ – cot θ
= R.H.S.
– Hence, proved.
12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1+ cos θ), we get
= R.H.S.
– Hence, proved.
13. sin θ/ (1 – cos θ) = cosec θ + cot θ
Solution:
Taking L.H.S.,
= cosec θ + cot θ
= R.H.S.
– Hence, proved.
RD Sharma Class 10 Solutions Chapter 6 PDF- Direct LiNK
Exercise-Wise RD Sharma Class 10 Maths Chapter 6 – Trigonometric Identities
| RD Sharma Class 10 Solutions Chapter 6 Exercises |
| RD Sharma Solutions for Class 10 Chapter 6 Exercise 6.1 |
| RD Sharma Solutions for Class 10 Chapter 6 Exercise 6.2 |
| RD Sharma Solutions for Class 10 Chapter 6 VSAQs |
| RD Sharma Solutions for Class 10 Chapter 6 MCQs |
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