$\frac{(s i n 7 x + s i n 5 x) + (s i n 9 x + s i n 3 x)}{(c o s 7 x + c o s 5 x) + (c o s 9 x + c o s 3 x)} = t a n 6 x$

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Solution

L.H.S $= \frac{(sin 7 x + sin 5 x) + (sin 9 x + sin 3 x)}{(cos 7 x + cos 5 x) + (cos 9 x + cos 3 x)}$
$= \frac{2 sin (\frac{7 x + 5 x}{2}) cos (\frac{7 x - 5 x}{2}) + 2 sin (\frac{9 x + 3 x}{2}) cos (\frac{9 x - 3 x}{2})}{2 cos (\frac{7 x + 5 x}{2}) cos (\frac{7 x - 5 x}{2}) + 2 cos (\frac{9 x + 3 x}{2}) cos (\frac{9 x - 3 x}{2})}$
$= \frac{sin (6 x) cos (x) + sin (6 x) cos (3 x)}{cos (6 x) cos (x) + cos (6 x) cos (3 x)}$
$= \frac{sin 6 x (cos x + cos 3 x)}{cos 6 x (cos x + cos 3 x)}$
$= \frac{sin 6 x}{cos 6 x} = tan 6 x =$ R.H.S
Hence proved.

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