Prove the following:
${sin}^{2} 6 x - {sin}^{2} 4 x = sin 2 x, sin 10 x$

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Solution

LHS $= {sin}^{2} 6 x - {sin}^{2} 4 x$

$= (sin 6 x + sin 4 x) (sin 6 x - sin 4 x)$ ... [ $∵ a^{2} - b^{2} = (a + b) (a - b)$ ]

We know that, $sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$

$∴$ LHS $= [2 sin (\frac{6 x + 4 x}{2}) cos (\frac{6 x - 4 x}{2})]$ $[2 cos (\frac{6 x + 4 x}{2}) sin (\frac{6 x - 4 x}{2})]$

$= [2 sin (\frac{10 x}{2}) cos (\frac{2 x}{2})]$ $[2 cos (\frac{10 x}{2}) sin (\frac{2 x}{2})]$

$= 2 sin 5 x cos x \times 2 cos 5 x sin x$

$= 2 sin 5 x cos 5 x \times 2 sin x cos x$

$= sin 10 x \times sin 2 x$ ... $[∵ sin 2 θ = 2 sin θ cos θ]$

$=$ RHS

Hence proved.

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Prove that: ${sin}^{2} 6 x - {sin}^{2} 4 x = sin 2 x sin 10 x$

sin² 6x – sin² 4x = sin 2x sin 10x

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