(sin 0 + sec )2 +(cos 0 + cosec 6)2
= (1 + sec 0 cosec 0)2
Answers
Answer:
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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
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1
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
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 2
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,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3
= cos2 A + sin2 A
= 1
= R.H.S
– Hence Proved
10.