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Question
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
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Solution
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
`(1 + cosA)/sinA + sinA/(1 + cosA)`
= `((1 + cosA)^2 + sin^2A)/(sinA(1 + cosA))`
= `(1 + 2cosA + cos^2A + sin^2A)/(sinA(1 + cosA))`
= `(2 + 2cosA)/(sinA(1 + cosA))`
= `(2(1 + cosA))/(sinA(1 + cosA)` [`sin^2A + cos^2A = 1`]
= 2 cosec A
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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
