Answer
$$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$$
The equation is an identity, as shown below.
Work Step by Step
$$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$$
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.
