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Prove $\cot(x) - \cot(2x) =\csc(2x)$. I start to solve from LHS, and change all the terms into $\sin$ and $\cos$, but I could not prove it into $\csc(2x)$.

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Notice, $$\cot x-\cot 2x$$ $$=\frac{1}{\tan x}-\frac{1}{\tan 2x}$$ $$=\frac{1}{\tan x}-\frac{1-\tan^2x}{2\tan x}$$ $$=\frac{2-1+\tan^2 x}{2\tan x}$$ $$=\frac{1+\tan^2 x}{2\tan x}$$ $$=\frac{\sec^2 x}{2\tan x}$$ $$=\frac{1}{2\tan x \cos^2 x}$$ $$=\frac{1}{2\sin x\cos x}$$ $$=\frac{1}{\sin 2x}$$ $$=\color{blue}{\csc 2x}$$

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We could use the difference identity for $\sin(\alpha - \beta)$. Alternatively (since I can never remember those identities), we could go the longer route and use double angle identities: \begin{align*} \cot x - \cot 2x &= \frac{\cos x}{\sin x} - \frac{\cos 2x}{\sin 2x} \\ &= \frac{\cos x}{\sin x} - \frac{2\cos^2 x - 1}{2\sin x \cos x} \\ &= \frac{(2\cos^2 x) - (2\cos^2 x - 1)}{2\sin x \cos x} \\ &= \frac{1}{2\sin x \cos x} \\ &= \frac{1}{\sin 2x} \\ &= \csc 2x \end{align*}

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$$\begin{align} \cot{x}-\cot{2x}&={2\cos^2{x}-\cos{2x}\over 2\sin{x}\cos{x}}\\ &={1+\cos{2x}-\cos{2x}\over 2\sin{x}\cos{x}}\\ &={1\over \sin{2x}}\\ &=\csc{2x} \end{align}$$

Where we have use the two identitities $\sin{2x}=2\sin{x}\cos{x}$ and $2\cos^2{x}-1=\cos{2x}$

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So we wish to show that $\cot{x}-\cot{2x}=\csc{2x}$. Rearranging the equation we can show $\cot{2x}+\csc{2x}=\cot{x}$.

Thus we have $$\cot{2x}=\frac{1}{\tan{2x}}=\frac{1-\tan^2{x}}{2\tan{x}} $$ and $$\csc{2x}=\frac{1}{\sin{2x}}=\frac{1}{2\sin{x}\cos{x}}\equiv \frac{\frac{1}{\cos^2{x}}}{\frac{2\sin{x}}{\cos{x}}}=\frac{\sec^2{x}}{2\tan{x}} $$ Thus via the identity $\sec^2{x}-\tan^2{x}=1 $ we have

$$\cot{2x}+\csc{2x}=\frac{1-\tan^2{x}}{2\tan{x}}+\frac{\sec^2{x}}{2\tan{x}} $$

$$=\frac{1+\sec^2{x}-\tan^2{x}}{2\tan{x}} $$

$$=\frac{2}{2\tan{x}} $$

$$=\cot{x} $$

(Via the identities there are a plethora of ways to solve it!)

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