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Question
Prove that `(tan(90 - θ) + cot(90 - θ))/("cosec" θ) = sec θ`.
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Solution
L.H.S. = `(tan(90 - θ) + cot(90 - θ))/("cosec" θ)`
= `1/("cosec" θ)(cot θ + tan θ)` ...`[(∵ tan(90 - θ) = cot θ),(cot(90 - θ) = tan θ)]`
= sin θ (cot θ + tan θ)
= `sin θ ((cos θ)/(sin θ) + (sin θ)/(cos θ))`
= `sin θ ((cos^2θ + sin^2θ)/(sinθ cosθ))`
= `sin θ (1/(sin θ cos θ))` ...[∵ sin2θ + cos2θ = 1]
= `1/(cos θ)`
= sec θ
= R.H.S.
∴ `(tan(90 - θ) + cot(90 - θ))/("cosec" θ) = sec θ`
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