Answer
$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$
The equation has been verified to be an identity as below.
Work Step by Step
$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$
We examine from the left side.
$$X=\frac{2}{1+\cos x}-\tan^2\frac{x}{2}$$
$$X=\frac{2}{1+\cos x}-\Big(\tan\frac{x}{2}\Big)^2$$
From the half-angle identity for tangent:
$$\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$$
Therefore, $$\Big(\tan\frac{x}{2}\Big)^2=\frac{\sin^2x}{(1+\cos x)^2}$$
Apply back to $X$:
$$X=\frac{2}{1+\cos x}-\frac{\sin^2x}{(1+\cos x)^2}$$
$$X=\frac{2(1+\cos x)-\sin^2 x}{(1+\cos x)^2}$$
$$X=\frac{2+2\cos x-\sin^2x}{(1+\cos x)^2}$$
We can rewrite $\sin^2x$ into $1-\cos^2x$ according to Pythagorean identities.
$$X=\frac{2+2\cos x-(1-\cos^2x)}{(1+\cos x)^2}$$
$$X=\frac{2+2\cos x-1+\cos^2x}{(1+\cos x)^2}$$
$$X=\frac{1+2\cos x+\cos^2x}{(1+\cos x)^2}$$
$$X=\frac{(1+\cos x)^2}{(1+\cos x)^2}$$
$$X=1$$
That concludes $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$
2 sides are equal, so the equation is an identity.
