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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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संबंधित प्रश्न
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
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