Answer
As proved before, the equation is an identity.
$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$
Work Step by Step
$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$
The left side would be examined first.
$$X=1-\tan^2\frac{\theta}{2}$$
- Half-angle identity for tangent: $\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ (we choose this one since here it involves $\tan^2\frac{\theta}{2}$)
Thus, $$\tan^2\frac{\theta}{2}=\frac{1-\cos\theta}{1+\cos\theta}$$ (After squaring, the $\pm$ sign can be removed)
Replace back to $X$:
$$X=1-\frac{1-\cos\theta}{1+\cos\theta}$$
$$X=\frac{1+\cos\theta-(1-\cos\theta)}{1+\cos\theta}$$
$$X=\frac{1+\cos\theta-1+\cos\theta}{1+\cos\theta}$$
$$X=\frac{2\cos\theta}{1+\cos\theta}$$
Therefore, $$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$
The equation is an identity.
