Question

Prove that $\frac{sin5x+sin3x}{sin\,7x+cos3x}=tan\,4x$

Answer 1

It is known that

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

$∴ L.H.S=\frac{sin5x+sin3x}{sin\,7x+cos3x}$

$=\frac{-2\,sin(\frac{5x+3x}{2}).cos(\frac{5x-3x}{2})}{2cos(\frac{5x+3x}{2}).cos(\frac{5x-3x}{2})}$

$=\frac{-2\,sin4x.cos2x}{2cos4x.cos7x}$

$=-\frac{sin4x}{cos4x}$

$=tan\,4x=R.H.S$