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Question
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
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Solution
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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