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प्रश्न
Prove that `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
उत्तर
L.H.S = `(sintheta + "cosec" theta)/sin theta`
= `sintheta/sintheta + ("cosec"theta)/sintheta`
= 1 + cosec θ × cosec θ ......`[∵ "cosec" theta = 1/sin theta]`
= 1 + cosec2θ
= 1 + 1 + cot2θ .......[∵ 1 + cot2θ = cosec2θ]
= 2 + cot2θ
= R.H.S
∴ `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
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= 12
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