RD Sharma Solutions For Class 10 Chapter 6 – Trigonometric Identities
RD Sharma Class 10 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,
= 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
14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ)2
Solution:
Taking the L.H.S,
= (sec θ – tan θ)2
= R.H.S
– Hence Proved
15. 
Solution:
Taking L.H.S,
= cot θ
= R.H.S
– Hence Proved
16. tan2 θ − sin2 θ = tan2 θ sin2 θ
Solution:
Taking L.H.S,
L.H.S = tan2 θ − sin2 θ
= tan2 θ sin2 θ
= R.H.S
– Hence Proved
17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θ
Solution:
Taking L.H.S = (cosec θ + sin θ)(cosec θ – sin θ)
On multiplying we get,
= cosec2 θ – sin2 θ
= (1 + cot2 θ) – (1 – cos2 θ) [Using cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1]
= 1 + cot2 θ – 1 + cos2 θ
= cot2 θ + cos2 θ
= R.H.S
– Hence Proved
18. (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ
Solution:
Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ)
On multiplying we get,
= sec2 θ – sin2 θ
= (1 + tan2 θ) – (1 – sin2 θ) [Using sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1]
= 1 + tan2 θ – 1 + sin2 θ
= tan 2 θ + sin 2 θ
= R.H.S
– Hence Proved
19. sec A(1- sin A) (sec A + tan A) = 1
Solution:
Taking L.H.S = sec A(1 – sin A)(sec A + tan A)
Substituting sec A = 1/cos A and tan A =sin A/cos A in the above we have,
L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A)
= 1 – sin2 A / cos2 A [After taking L.C.M]
= cos2 A / cos2 A [∵ 1 – sin2 A = cos2 A]
= 1
= R.H.S
– Hence Proved
20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Solution:
Taking L.H.S = (cosec A – sin A)(sec A – cos A)(tan A + cot A)
= (cos2 A/ sin A) (sin2 A/ cos A) (1/ sin A cos A) [∵ sin2 θ + cos2 θ = 1]
= sin A x cos A x (1/ cos A sin A)
= R.H.S
– Hence Proved
21. (1 + tan2 θ)(1 – sin θ)(1 + sin θ) = 1
Solution:
Taking L.H.S = (1 + tan2θ)(1 – sin θ)(1 + sin θ)
And, we know sin2 θ + cos2 θ = 1 and sec2 θ – tan2 θ = 1
So,
L.H.S = (1 + tan2 θ)(1 – sin θ)(1 + sin θ)
= (1 + tan2 θ){(1 – sin θ)(1 + sin θ)}
= (1 + tan2 θ)(1 – sin2 θ)
= sec2 θ (cos2 θ)
= (1/ cos2 θ) x cos2 θ
= 1
= R.H.S
– Hence Proved
22. sin2 A cot2 A + cos2 A tan2 A = 1
Solution:
We know that,
cot2 A = cos2 A/ sin2 A and tan2 A = sin2 A/cos2 A
Substituting the above in L.H.S, we get
L.H.S = sin2 A cot2 A + cos2 A tan2 A
= {sin2 A (cos2 A/ sin2 A)} + {cos2 A (sin2 A/cos2 A)}
= cos2 A + sin2 A
= 1 [∵ sin2 θ + cos2 θ = 1]
= R.H.S
– Hence Proved
23.
Solution:
(i) Taking the L.H.S and using sin2 θ + cos2 θ = 1, we have
L.H.S = cot θ – tan θ
= R.H.S
– Hence Proved
(ii) Taking the L.H.S and using sin2 θ + cos2 θ = 1, we have
L.H.S = tan θ – cot θ
= R.H.S
– Hence Proved
24. (cos2 θ/ sin θ) – cosec θ + sin θ = 0
Solution:
Taking L.H.S and using sin2 θ + cos2 θ = 1, we have
= – sin θ + sin θ
= 0
= R.H.S
- Hence proved
25.
Solution:
Taking L.H.S,
= 2 sec2 A
= R.H.S
- Hence proved
26. 
Solution:
Taking the LHS and using sin2 θ + cos2 θ = 1, we have
= 2/ cos θ
= 2 sec θ
= R.H.S
- Hence proved
27.
Solution:
Taking the LHS and using sin2 θ + cos2 θ = 1, we have
= R.H.S
- Hence proved
28.
Solution:
Taking L.H.S,
Using sec2 θ − tan2 θ = 1 and cosec2 θ − cot2 θ = 1
= R.H.S
- Hence proved
29.
Solution:
Taking L.H.S and using sin2 θ + cos2 θ = 1, we have
= R.H.S
- Hence proved
30.
Solution:
Taking LHS, we have
= 1 + tan θ + cot θ
= R.H.S
- Hence proved
31. sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Solution:
From trig. Identities we have,
sec2 θ − tan2 θ = 1
On cubing both sides,
(sec2θ − tan2θ)3 = 1
sec6 θ − tan6 θ − 3sec2 θ tan2 θ(sec2 θ − tan2 θ) = 1
[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]sec6 θ − tan6 θ − 3sec2 θ tan2 θ = 1
⇒ sec6 θ = tan6 θ + 3sec2 θ tan2 θ + 1
Hence, L.H.S = R.H.S
- Hence proved
32. cosec6 θ = cot6 θ + 3cot2 θ cosec2 θ + 1
Solution:
From trig. Identities we have,
cosec2 θ − cot2 θ = 1
On cubing both sides,
(cosec2 θ − cot2 θ)3 = 1
cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ (cosec2 θ − cot2 θ) = 1
[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ = 1
⇒ cosec6 θ = cot6 θ + 3 cosec2 θ cot2 θ + 1
Hence, L.H.S = R.H.S
- Hence proved
33.
Solution:
Taking L.H.S and using sec2 θ − tan2 θ = 1 ⇒ 1 + tan2 θ = sec2 θ
= R.H.S
- Hence proved
34.
Solution:
Taking L.H.S and using the identity sin2A + cos2A = 1, we get
sin2A = 1 − cos2A
⇒ sin2A = (1 – cos A)(1 + cos A)
- Hence proved
35.
Solution:
We have,
Rationalizing the denominator and numerator with (sec A + tan A) and using sec2 θ − tan2 θ = 1 we get,
= R.H.S
- Hence proved
36.
Solution:
We have,
On multiplying numerator and denominator by (1 – cos A), we get
= R.H.S
- Hence proved
37. (i)
Solution:
Taking L.H.S and rationalizing the numerator and denominator with √(1 + sin A), we get
= R.H.S
- Hence proved
(ii)
Solution:
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get
= 2 cosec A
= R.H.S
- Hence proved
38. Prove that:
(i)
Solution:
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get
= R.H.S
- Hence proved
(ii)
Solution:
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get
= R.H.S
- Hence proved
(iii)
Solution:
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get
= 2 cosec θ
= R.H.S
- Hence proved
(iv)
Solution:
Taking L.H.S, we have
= R.H.S
- Hence proved
39.
Solution:
Taking LHS = (sec A – tan A)2 , we have
= R.H.S
- Hence proved
40.
Solution:
Taking L.H.S and rationalizing the numerator and denominator with (1 – cos A), we get
= (cosec A – cot A)2
= (cot A – cosec)2
= R.H.S
- Hence proved
41.
Solution:
Considering L.H.S and taking L.C.M and on simplifying we have,
= 2 cosec A cot A = RHS
- Hence proved
42.
Solution:
Taking LHS, we have
= cos A + sin A
= RHS
- Hence proved
43.
Solution:
Considering L.H.S and taking L.C.M and on simplifying we have,
= 2 sec2 A
= RHS
- Hence proved
RD Sharma Class 10 Chapter 6 Exercise 6.2 Page No: 6.54
1. If cos θ = 4/5, find all other trigonometric ratios of angle θ.
Solution:
We have,
cos θ = 4/5
And we know that,
sin θ = √(1 – cos2 θ)
⇒ sin θ = √(1 – (4/5)2)
= √(1 – (16/25))
= √[(25 – 16)/25]
= √(9/25)
= 3/5
∴ sin θ = 3/5
Since, cosec θ = 1/ sin θ
= 1/ (3/5)
⇒ cosec θ = 5/3
And, sec θ = 1/ cos θ
= 1/ (4/5)
⇒ cosec θ = 5/4
Now,
tan θ = sin θ/ cos θ
= (3/5)/ (4/5)
⇒ tan θ = 3/4
And, cot θ = 1/ tan θ
= 1/ (3/4)
⇒ cot θ = 4/3
2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.
Solution:
We have,
sin θ = 1/√2
And we know that,
cos θ = √(1 – sin2 θ)
⇒ cos θ = √(1 – (1/√2)2)
= √(1 – (1/2))
= √[(2 – 1)/2]
= √(1/2)
= 1/√2
∴ cos θ = 1/√2
Since, cosec θ = 1/ sin θ
= 1/ (1/√2)
⇒ cosec θ = √2
And, sec θ = 1/ cos θ
= 1/ (1/√2)
⇒ cosec θ = √2
Now,
tan θ = sin θ/ cos θ
= (1/√2)/ (1/√2)
⇒ tan θ = 1
And, cot θ = 1/ tan θ
= 1/ (1)
⇒ cot θ = 1
3.
Solution:
Given,
tan θ = 1/√2
By using sec2 θ − tan2 θ = 1,
4.
Solution:
Given,
tan θ = 3/4
By using sec2 θ − tan2 θ = 1,
sec θ = 5/4
Since, sec θ = 1/ cos θ
⇒ cos θ = 1/ sec θ
= 1/ (5/4)
= 4/5
So,
5.
Solution:
Given, tan θ = 12/5
Since, cot θ = 1/ tan θ = 1/ (12/5) = 5/12
Now, by using cosec2 θ − cot2 θ = 1
cosec θ = √(1 + cot2 θ)
= √(1 + (5/12)2 )
= √(1 + 25/144)
= √(169/ 25)
⇒ cosec θ = 13/5
Now, we know that
sin θ = 1/ cosec θ
= 1/ (13/5)
⇒ sin θ = 5/13
Putting value of sin θ in the expression we have,
= 25/ 1
= 25
6.
Solution:
Given,
cot θ = 1/√3
Using cosec2 θ − cot2 θ = 1, we can find cosec θ
cosec θ = √(1 + cot2 θ)
= √(1 + (1/√3)2)
= √(1 + (1/3)) = √((3 + 1)/3)
= √(4/3)
⇒ cosec θ = 2/√3
So, sin θ = 1/ cosec θ = 1/ (2/√3)
⇒ sin θ = √3/2
And, we know that
cos θ = √(1 – sin2 θ)
= √(1 – (√3/2)2)
= √(1 – (3/4))
= √((4 – 3)/4)
= √(1/4)
⇒ cos θ = 1/2
Now, using cos θ and sin θ in the expression, we have
= 3/5
7.
Solution:
Given,
cosec A = √2
Using cosec2 A − cot2 A = 1, we find cot A
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