## Trigonometry (11th Edition) Clone

$$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$$ The equation is an identity, as shown below.
$$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$$ The left side, which is more complex, should be begun with. $$X=\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}$$ - For $\tan\frac{x}{2}$: Apply the half-angle identity for tangent, we have $$\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$$ - For $\cot\frac{x}{2}$: We already know from reciprocal identities that $\cot\frac{x}{2}=\frac{1}{\tan\frac{x}{2}}$ However, for more convenience in calculations later, we would not use the identity for $\tan\frac{x}{2}$ above, but this one $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$. So, $$\cot\frac{x}{2}=\frac{1}{\tan\frac{x}{2}}=\frac{1}{\frac{\sin x}{1+\cos x}}=\frac{1+\cos x}{\sin x}$$ Therefore, $X$ would be $$X=\frac{1}{2}\times\frac{1+\cos x}{\sin x}-\frac{1}{2}\times\frac{1-\cos x}{\sin x}$$ $$X=\frac{1+\cos x}{2\sin x}-\frac{1-\cos x}{2\sin x}$$ $$X=\frac{1+\cos x-1+\cos x}{2\sin x}$$ $$X=\frac{2\cos x}{2\sin x}$$ $$X=\frac{\cos x}{\sin x}$$ - Finally, from quotient identities: $\frac{\cos x}{\sin x}=\cot x$. Thus, $$X=\cot x$$ That means $$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$$ The equation is an identity.