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प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove that `(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
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उत्तर
L.H.S = `(sin theta-2sin^3theta)/(2cos^3theta -costheta)`
= `(sintheta(1-2sin^2theta))/(costheta(2cos^2theta-1))`
= `(sinthetaxx(1-2sin^2theta))/(costhetaxx{2(1-sin^2theta)-1})`
= `(sin thetaxx(1-2sin^2theta))/(costhetaxx(1-2sin^2theta))`
= `tantheta`
= R.H.S
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
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L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
