Question

Prove that $\frac{(sin 7x+sin 5x)+(sin 9x+sin3x)}{cos 7x+cos5x)+(cos9x+cos3x)}=tan\,6x.$

Answer 1

It is known that

$sinA+sin B=2sin(\frac{A+B}{2}).cos(\frac{A-B}{2}),cosA+cosB=2cos(\frac{A+B}{2}).cos(\frac{A-B}{2})$

L.H.S. = $\frac{(sin 7x+sin 5x)+(sin 9x+sin3x)}{(cos 7x+cos5x)+(cos9x+cos3x)}$

$\frac{[2sin(\frac{7x+5x}{2}).cos(\frac{7x-5x}{2})]+[2sin(\frac{9x+3x}{2}.cos(\frac{9x-3x}{2})]}{[2cos(\frac{7x+5x}{2}.cos(\frac{7x-5x}{2})]+[2cos(\frac{9x+3x}{2}).cos(\frac{9x-3x}{2})]}$

$=\frac{[2sin6x.cosx]+[2sin6x.cos3x]}{[2cos6x.cosx]+[2cos6x.cos3x]}$

$=\frac{2sin6x[cosx+cos3x]}{2cos6x[cosx+cos3x]}$

$= tan\, 6x$

$= R.H.S.$