## Trigonometry (11th Edition) Clone

As proved before, the equation is an identity. $$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$
$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ 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.