Answer
The equation $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ is an identity, as 2 sides are proved to be equal below.
Work Step by Step
$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$
The left side would be examined first.
$$X=\frac{2}{1+\cos x}-\tan^2\frac{x}{2}$$
- Half-angle identity for tangent:
$$\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$$ (we choose this identity because here we would use $\tan^2\frac{x}{2}$)
$$\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$
(The $\pm$ sign can be removed since both $\tan^2\frac{x}{2}$ and $\frac{1-\cos x}{1+\cos x}$ are positive)
Apply back to $X$, we have
$$X=\frac{2}{1+\cos x}-\frac{1-\cos x}{1+\cos x}$$
$$X=\frac{2-(1-\cos x)}{1+\cos x}$$
$$X=\frac{2-1+\cos x}{1+\cos x}$$
$$X=\frac{1+\cos x}{1+\cos x}$$
$$X=1$$
So the left side is equal to the right side. The equation $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ is therefore an identity.
