Question

sin5x+sin3x/cos5x+cos3x=tan4x  prove that

Answer 1

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

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

$\frac{sin5x+sin3x}{cos5x+cos3x}= \frac{2*sin( \frac{5x+3x}{2} )*cos( \frac{5x-3x}{2} )}{2*cos( \frac{5x+3x}{2} )*cos( \frac{5x-3x}{2} )}$

$\frac{2*sin4x*cosx}{2*cos4x*cosx} = \frac{sin4x}{cos4x} =tan4x$