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Question
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
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Solution
L.H.S. = `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1)`
= `(sin θ)/(1/cos θ + 1) + (sin θ)/(1/(cos θ) - 1`
= `(sin θ)/((1 + cos θ)/(cos θ)) + (sin θ)/((1 - cos θ)/(cos θ))`
= `(sin θ cos θ)/(1 + cos θ) + (sin θ cos θ)/(1 - cos θ)`
= `sin θ cos θ (1 /(1 + cos θ) + 1/(1 - cos θ))`
= `sin θ cos θ [(1 - cos θ + 1 + cos θ)/((1 + cos θ)(1 - cos θ))]`
= `sin θ cos θ (2/(1 - cos^2θ))` ...[∵ (a + b)(a – b) = a2 – b2]
= `sin θ cos θ xx 2/(sin^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
= `2 xx (cos θ)/(sin θ)`
= 2 cot θ
= R.H.S.
∴ `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`
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