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प्रश्न
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
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उत्तर
LHS = `(cotA - cosecA)^2`
= `[cosA/sinA - 1/sinA]^2`
= `[(cosA - 1)/sinA]^2`
= `(cosA - 1)^2/sin^2A = (cosA - 1)^2/(1 - cos^2A)`
= `(-(1 - cosA))^2/((1 - cosA)(1 + cosA)) = ((1 - cosA)(1 - cosA))/((1 - cosA)(1 + cosA))`
= `(1 - cosA)/(1 + cosA)`
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Show that, cotθ + tanθ = cosecθ × secθ
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
