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Question
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
Solution
L.H.S = `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2`
= `(1 + sin^2theta + cos^2theta + 2sintheta - 2sintheta cos theta - 2costheta)/(1 + sin^2theta + cos^2theta + 2sintheta + 2sintheta costheta + 2costheta)`
= `(1 + 1 + 2sintheta (1 - cos theta) - 2cos theta)/(1 + 1 + 2sin theta + 2cos theta (sin theta + 1))`
= `(2(1 - cos theta) + 2sintheta (1 - cos theta))/(2(1 + sin theta) + 2cos theta(1 + sin theta))`
= `(2(1 - costheta)(1 + sintheta))/(2(1 + sintheta)(1 + costheta))`
= `((1 - cos theta))/((1 + cos theta))`
L.H.S = R.H.S
Hence it is proved.
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