Prove that : sinA- 2sin^A / …

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Sia ? 6 years, 4 months ago
LHS = {tex}\frac{\sin A-2 \sin ^{3} A}{2 \cos ^{3} A-\cos A}{/tex}
= {tex}\frac{\sin A\left(1-2 \sin ^{2} A\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex}
= {tex}\frac{\sin A\left(1-2\left(1-\cos ^{2} A\right)\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex} [{tex}\because sin^2\theta=1-cos^2\theta{/tex}]
= {tex}\frac{\sin A\left(1-2+2cos ^{2} A\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex}
= {tex}\tan A \frac{\left(2 \cos ^{2} A-1\right)}{\left(2 \cos ^{2} A-1\right)}{/tex}
= tan A = RHS
Hence proved.
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