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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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