Answer
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
The equation is an identity.
Work Step by Step
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
We would examine the right side first, since it is more complex.
$$X=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
- Half-angle identity for tangent: $\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$
Therefore, $$\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$
(As 2 sides are both positive, we do not need to use the $\pm$ sign anymore)
Apply the identity back to $X$:
$$X=\frac{1-\frac{1-\cos x}{1+\cos x}}{1+\frac{1-\cos x}{1+\cos x}}$$
$$X=\frac{\frac{1+\cos x-(1-\cos x)}{1+\cos x}}{\frac{1+\cos x+1-\cos x}{1+\cos x}}$$
$$X=\frac{\frac{1+\cos x-1+\cos x}{1+\cos x}}{\frac{2}{1+\cos x}}$$
$$X=\frac{\frac{2\cos x}{1+\cos x}}{\frac{2}{1+\cos x}}$$
$$X=\frac{2\cos x}{2}$$
$$X=\cos x$$
As a result, 2 sides are equal and the equation is verified to be an identity.
