## Trigonometry (11th Edition) Clone

$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ The equation has been verified to be an identity as below.
$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ We examine from the left side. $$X=\frac{2}{1+\cos x}-\tan^2\frac{x}{2}$$ $$X=\frac{2}{1+\cos x}-\Big(\tan\frac{x}{2}\Big)^2$$ From the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$$ Therefore, $$\Big(\tan\frac{x}{2}\Big)^2=\frac{\sin^2x}{(1+\cos x)^2}$$ Apply back to $X$: $$X=\frac{2}{1+\cos x}-\frac{\sin^2x}{(1+\cos x)^2}$$ $$X=\frac{2(1+\cos x)-\sin^2 x}{(1+\cos x)^2}$$ $$X=\frac{2+2\cos x-\sin^2x}{(1+\cos x)^2}$$ We can rewrite $\sin^2x$ into $1-\cos^2x$ according to Pythagorean identities. $$X=\frac{2+2\cos x-(1-\cos^2x)}{(1+\cos x)^2}$$ $$X=\frac{2+2\cos x-1+\cos^2x}{(1+\cos x)^2}$$ $$X=\frac{1+2\cos x+\cos^2x}{(1+\cos x)^2}$$ $$X=\frac{(1+\cos x)^2}{(1+\cos x)^2}$$ $$X=1$$ That concludes $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ 2 sides are equal, so the equation is an identity.