Question

$if \: tan \: a = cos2 \alpha \: prove \: that \: \: \\ sin2a = \frac{1 - {tan}^{4} \alpha }{1 + {tan}^{4} \alpha }$
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Answer 1

Trigonometry... just apply rules for expansion of cos 2A and sin2A in terms of sinA and cosA,,,

\frac{sec8x-1}{sec4x-1}=\frac{(1-cos8x)cos4x}{cos8x(1-cos4x)}\\\\=\frac{2sin^24x\ cos4x}{cos8x\ *\ 2\ sin^2 2x}=\frac{2sin4x\ cos4x\ *sin4x}{cos8x\ *sin^22x}\\\\=\frac{sin8x*2sin2x\ cos2x}{cos8x\ * sin^22x}=\frac{tan8x}{tan2x}

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