Prove the following trigonometric identities.
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Answer with Step-by-step explanation:
Given :
(1+ sinθ)/ cosθ + cosθ /(1 + sinθ) = 2secθ
LHS = (1+ sinθ)/ cosθ + cosθ /(1 + sinθ)
= (1 + sinθ)² + cos²θ) / [cosθ(1+sinθ)]
[By taking LCM]
= (1 + sin²θ + 2sinθ + cos²θ)/[cosθ(1+sinθ)]
[By using identity , (a + b)² = a² + 2ab + b²]
= ( sin²θ +cos²θ + 2sinθ + 1)/[cosθ(1+sinθ)]
= ( 1 + 2sinθ + 1)/[cosθ(1+sinθ)]
[By using the identity , sin² θ + cos² θ = 1]
= ( 2 + 2sinθ )/[cosθ(1+sinθ)]
= 2(1 + sinθ)/ [cosθ(1+sinθ)]
= 2/cosθ
= 2 × 1/cosθ
= 2 secθ
[By using the identity, secθ = 1/ cosθ]
L.H.S = R.H.S
Hence Proved..
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