Answer
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
The proof is explained in the work step by step below.
Work Step by Step
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
We examine the right side first.
$$X=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
- Half-angle identity for tangent: $$\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$$
So it follows that
$$\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$
Apply the identity to $X$
$$X=\frac{1-\frac{1-\cos x}{1+\cos x}}{1+\frac{1-\cos x}{1+\cos x}}$$
$$X=\frac{\frac{1+\cos x-1+\cos x}{1+\cos x}}{\frac{1+\cos x+1-\cos x}{1+\cos x}}$$
$$X=\frac{1+\cos x-1+\cos x}{1+\cos x+1-\cos x}$$
$$X=\frac{2\cos x}{2}$$
$$X=\cos x$$
Thus, $$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
The equation has been verified to be an identity.
