Trigonometry (11th Edition) Clone

$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ The proof is explained in the work step by step below.
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ 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.