Question

sin4x + sin2x/1+ cos4x+cos2x = tan2x​

Answer 1

starting with L.H.S we can write the numerator

sin2x + sin4x = sin2x + 2 sin2xcos2x= sin2x(1+2cos2x)

since sin 2 theta = 2 sin theta cos theta

implies sin 4x = 2 sin 2x cos 2x

for the the denominator

1+cos 2x + cos 4x ,

using cos 2 theta = cos^{2} theta- sin^{2} theta = 2cos^{2} theta -1

we substitute cos 4x as 2cos^{2} 2x -1

so 1+cos 2x + cos 4x = 1+cos 2x + 2cos^{2} 2x - 1 =

cos 2x + 2cos^{2} 2x i.e cos2x(1+2cos2x)

so the numerator and denominator now have a common factor as

(1+2cos2x)