Prove that: x - 3x ^2 x - ^2 x = 2 x - Vidyadip

Question

Prove that: $\frac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x$

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Answer

$\style{font-family:'Times New Roman'}{\style{font-size:8px}{\begin{array}{l}We\;have,\\L.H.S\;=\;\frac{\sin x-\sin3x}{\sin^{{}^2}x-\cos^{{}^2}x}=\;\frac{-(\sin3x-\sin x)}{-(\cos^{{}^2}x-\mathrm{si}n^2x)}\;=\;\frac{(\sin3x-\sin x)}{(\cos^2x-\sin^{{}^2}x)}\\\;\;\;\;\;\;\;=\frac{2\cos\left(\frac{3x+x}2\right)\sin\left(\frac{3x-x}2\right)}{\cos2x}\\\;\;\;\;\;\;=\frac{2\cos2x\sin x}{\cos2x}=\;2\sin x\\\;\;\;\;\;\;\;\;\;\\\\\\\end{array}}}$

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