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Question
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
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Solution
L.H.S. = `(sinthetasin(90^@-theta))/cot(90^@-theta)`
= `(sinthetacostheta)/tantheta`
= `(sinthetacostheta)/(sintheta/costheta)`
= cos2θ
= 1 – sin2θ = R.H.S.
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