To Prove: \(\operatorname{cosec} 2 x+\cot 2 x=\cot x\)
Taking LHS,
\(=\operatorname{cosec} 2 x+\cot 2 x \ldots\) (i)
\(\operatorname{cosec} x=\frac{1}{\sin x} \& \cot x=\frac{\cos x}{\sin x}\)
Replacing \(x\) by \(2 x\), we get
$$
\operatorname{cosec} 2 x=\frac{1}{\sin 2 x} \& \cot 2 x=\frac{\cos 2 x}{\sin 2 x}
$$
So, eq. (i) becomes
$$
\begin{aligned}
&=\frac{1}{\sin 2 x}+\frac{\cos 2 x}{\sin 2 x} \\
&=\frac{1+\cos 2 x}{\sin 2 x} \\
&=\frac{2 \cos ^{2} x}{\sin 2 x}\left[\because 1+\cos 2 x=2 \cos ^{2} x\right] \\
&=\frac{2 \cos ^{2} x}{2 \sin x \cos x}[\because \sin 2 x=2 \sin x \cos x] \\
&=\frac{\cos x}{\sin x} \\
&=\cot x\left[\because \cot x=\frac{\cos x}{\sin x}\right] \\
&=\text { RHS }
\end{aligned}
$$
Hence Proved
