## Trigonometry (11th Edition) Clone

$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ The equation is an identity.
$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ 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.