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प्रश्न
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
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उत्तर
`(secθ - tanθ)^2`
= `(1/cosθ - sinθ/cosθ)^2`
= `((1 - sinθ)/cosθ)^2 = (1 - sinθ)^2/cos^2θ`
= `(1 - sinθ)^2/(1 - sin^2θ) = (1 - sinθ)^2/((1 -sinθ)(1 + sinθ))` (∵ `1 - sin^2θ = cos^2θ`
= `(1 - sinθ)/(1 + sinθ)`
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
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= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
