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Maharashtra State BoardSSC (English Medium) 10th Standard

Prove that (tan(90 – θ) + cot(90 – θ))/(cosec θ) = sec θ.

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Question

Prove that `(tan(90 - θ) + cot(90 - θ))/("cosec"  θ) = sec θ`.

Theorem
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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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Chapter 6: Trigonometry - Q.3 (B)

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