Prove: $\frac{(sin 7 x + sin 5 x) + (sin 9 x + sin 3 x)}{(cos 7 x cos 5 x) + (cos 9 x cos 3 x)} = tan 6 x$

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Solution

Taking LHS-

Solving numerator
$(sin 7 x + sin 5 x) + (sin 9 x + sin 3 x)$

$s i n x + s i n y = 2 sin (\frac{x + y}{2}) cos (\frac{x - y}{2})$

$\Rightarrow 2 sin (6 x) . cos (x) + 2 sin 6 x . cos 3 x$

$\Rightarrow 2 sin 6 x (c o s x + cos 3 x) \to (1)$

Now, solving denominator-

$(cos 7 x + cos 5 x) + (cos 9 x + cos 3 x)$

$c o s x + c o s y = 2 cos (\frac{x + y}{2}) cos (\frac{x - y}{2})$

$\Rightarrow 2 cos (6 x) cos x + 2 cos 6 x . cos 3 x$

$\Rightarrow 2 cos 6 x (cos x + cos 3 x) \to (2)$

Now solving LHS-

LHS $= \frac{(1)}{(2)} = \frac{sin 6 x}{cos 6 x} = tan 6 x =$ RHS

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