Prove that $\frac{sinA-2sin^{3}A}{2cos^{3}A-cosA}=tanA$.

Given: $\frac{sinA-2sin^{3}A}{2cos^{3}A-cosA}=tanA$.

To do: To prove the that $L.H.S.=R.H.S.$

Solution:  

$L.H.S.=\frac{sinA-2sin^{3}A}{2cos^{3}A-cosA}$

            $=\frac{sinA}{cosA}.\frac{( 1-2sin^{2}A)}{( 2cos^{2}A-1)}$

            $=\frac{sinA}{cosA}.\frac{( sin^{2}A+cos^{2}A-2sin^{2}A)}{( 2cos^{2}A-ssin^{2}A-cos^{2}A)}$                                   $( \because sin^{2}A+cos^{2}A=1)$

            $=\frac{sinA}{cosA}.\frac{( cos^{2}A-sin^{2}A)}{( cos^{2}A-sin^{2}A)}$

            $=\frac{sinA}{cosA}.1$

            $=tanA$                   $( \because \frac{sinA}{cosA}=tanA)$

            $=R.H.S.$

Hence proved that $\frac{sinA-2sin^{3}A}{2cos^{3}A-cosA}=tanA$.

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Updated on: 10-Oct-2022

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