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Question
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
Solution
L.H.S = `sintheta/(1 + cos theta) + (1 + cos theta)/sintheta`
Taking the L.C.M of the denominators,
We get,
= `(sin^2theta + (1 + cos theta)^2)/((1 + cos theta)* sintheta)`
= `(sin^2theta + 1 + cos^2theta + 2costheta)/((1 + costheta) * sin theta)`
Since, sin2θ + cos2θ = 1
= `(1 + 1 + 2costheta)/((1 + costheta) * sin theta)`
= `(2 + 2 cos theta)/((1 + cos theta) * sin theta)`
= `(2(1 + cos theta))/((1 + cos theta) * sin theta)`
Since, `1/sin theta` = cosec θ
= `2/sin theta`
= 2 cosec θ
R.H.S
Hence proved.
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