How to prove
$\frac{\sin^2(x)}{1+\cos(2x)} = \frac1{2} \tan^2(x)$
How to prove
$\frac{\sin^2(x)}{1+\cos(2x)} = \frac1{2} \tan^2(x)$
Hint: $1+ \cos{2x} = 1 + 2\cos^{2}{x} -1 = 2\cos^{2}{x}$
Hint: use the Pythagorean trigonometric identity $\sin^2 x+\cos^2 x=1$ and the double-angle formula $\cos 2x=\cos^2 x-\sin^2 x$
Do you know the addition formula $\cos(a+b)=\cos (a)\cos b−\sin (a)\sin (b)$ ? For $a=b=x$ yields $\cos(2x)=\cos^2 (x)−\sin^2 (x)$. Then insert this result into your identity.