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Question
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
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Solution
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
LHS = `(cotA + cosecA - 1)/(cotA - cosecA + 1)`
we know that , cosec2A - cot2A = 1
substituting this in the numerator
`(cosecA + cotA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` .....(x²-y²= (x+y)(x-y))
`(cosecA + cotA - (cosecA + cotA)(cosecA - cotA))/(cotA - cosecA + 1)`
taking common
`((cosec A + cot A)(1-cosec A + cot A) )/ (cot A - cosec A + 1)`
cancelling like terms in numerator and denominator
we are left with cosec A + cot A
`= 1/sin A + cos A/sin A`
`= (1+cos A) / sin A`
= RHS
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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
