## Trigonometry (11th Edition) Clone

$$\cot^2\frac{x}{2}=\frac{(1+\cos x)^2}{\sin^2 x}$$ The equation is verified to be an identity.
$$\cot^2\frac{x}{2}=\frac{(1+\cos x)^2}{\sin^2 x}$$ We start with the left side $$A=\cot^2\frac{x}{2}$$ $$A=\frac{\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}$$ We know that $\cos2x=2\cos^2x-1$, which means $\cos^2x=\frac{\cos 2x+1}{2}$. Therefore, $\cos^2\frac{x}{2}$ can also be written in the same way: $\cos^2\frac{x}{2}=\frac{\cos x+1}{2}$ Similarly, $\cos2x=1-2\sin^2 x$, which means, $\sin^2 x=\frac{1-\cos 2x}{2}$. That means, $\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$ Therefore, $$A=\frac{\frac{\cos x+1}{2}}{\frac{1-\cos x}{2}}$$ $$A=\frac{\cos x+1}{1-\cos x}$$ Now we multiply both numerator and denominator by $(\cos x+1)$. The reason is that on the right side, there is $(1+\cos x)^2$, but the left side now only has $(1+\cos x)$ $$A=\frac{\cos x+1}{1-\cos x}\frac{\cos x+1}{\cos x+1}$$ $$A=\frac{(1+\cos x)^2}{1-\cos^2 x}$$ $$A=\frac{(1+\cos x)^2}{\sin^2 x}$$ Both sides are hence equal. The equation has been proved to be an identity.