Prove the following trigonometric identities. 
Answers
Answer with Step-by-step explanation:
Given :
(1/sec²θ − cos²θ + 1/cosec²θ − sin²θ) sin²θ cos²θ = 1− sin²θ cos²θ / (2 + sin²θcos²θ)
L.H.S :
(1/sec²θ − cos²θ + 1/cosec²θ − sin²θ) sin²θ cos²θ
= [1/{(1/cos²θ) − cos²θ} + 1 / {(1/sin²θ) − sin²θ}] sin²θ cos²θ
[By using, secθ = 1/ cosθ & cosecθ = 1/sinθ]
= [1/{(1 − cos⁴θ)/ cos²θ} + 1 /{(1− sin⁴θ)/sin²θ}] sin²θ cos²θ
[By taking LCM]
= [(cos²θ/1 − cos⁴θ ) + (sin²θ / 1− sin⁴θ)] sin²θ cos²θ
= [{(cos²θ/(1 + cos²θ)(1 − cos²θ)} + (sin²θ / (1+ sin²θ) (1 - sin²θ)} ] sin²θ cos²θ
[By using identity , a² - b² = (a + b) (a - b) ]
= [{(cos²θ/(1 + cos²θ)(sin²θ)} + (sin²θ / (1+ sin²θ) (cos²θ)} ] sin²θ cos²θ
[By using an identity, (1- cos²θ) = sin²θ & (1- sin²θ) = cos²θ]
= [{(cos²θ × (1+ sin²θ) (cos²θ) + (sin²θ ×(1 + cos²θ)(sin²θ)} / sin²θ cos²θ (1 + cos²θ)(1 + sin²θ)] × sin²θ cos²θ
[By taking LCM]
= [{(cos⁴θ × (1 + sin²θ) + (sin⁴θ ×(1 + cos²θ)} / (1 + cos²θ)(1 + sin²θ)]
= [{(cos⁴θ + cos⁴θ sin²θ) + (sin⁴θ + sin⁴θ cos²θ)} / (1 + cos²θ + sin²θ + sin²θ cos²θ)]
= [{(cos⁴θ + sin⁴θ + cos²θ sin²θ(sin²θ + cos²θ)} / (1 + 1 + sin²θ cos²θ)]
[By using the identity , sin² θ + cos² θ = 1 & taking cos²θ sin²θ common]
= [{(cos⁴θ + sin⁴θ + cos²θ sin²θ(1)} / (2 + sin²θ cos²θ)]
[By using the identity , sin² θ + cos² θ = 1]
= [{(cos²θ)² + (sin²θ)² + cos²θ sin²θ} / (2 + sin²θ cos²θ)]
= [{(cos²θ + sin²θ)² - 2 cos²θ sin²θ + cos²θ sin²θ } / (2 + sin²θ cos²θ)]
[By using identity , a² + b² = (a + b)² - 2ab ]
= [{(1)² - cos²θ sin²θ } / (2 + sin²θ cos²θ)]
[By using the identity , sin² θ + cos² θ = 1]
= (1 - sin²θ cos²θ) / (2 + sin²θ cos²θ)]
(1/sec²θ − cos²θ + 1/cosec²θ − sin²θ) sin²θ cos²θ = 1− sin²θ cos²θ / (2 + sin²θcos²θ)
L.H.S = R.H.S
Hence Proved..
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