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प्रश्न
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
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उत्तर
L.H.S. = `(sin θ + "cosec" θ)/(sin θ)`
= `(sin θ)/(sin θ) + ("cosec" θ)/(sin θ)`
= 1 + cosec θ × cosec θ ...`[∵ "cosec" θ = 1/(sin θ)]`
= 1 + cosec2θ
= 1 + 1 + cot2θ ...[∵ 1 + cot2θ = cosec2θ]
= 2 + cot2θ
= R.H.S.
∴ `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
