Prove that \(\mathrm{sech}^{-1}(\sin\theta)=\log\left(\cot\frac{\theta}{2}\right)\).

Let \(\mathrm{sech}^{-1}(\sin\theta)=u\). Then \(\mathrm{sech}\,u=\sin\theta\Rightarrow\cosh u=\frac{1}{\sin\theta}\) (take \(0<\theta<\pi\)).

\(\sinh u=\sqrt{\cosh^2u-1}=\sqrt{\frac{1-\sin^2\theta}{\sin^2\theta}}=\frac{\cos\theta}{\sin\theta}\).

\(e^u=\cosh u+\sinh u=\frac{1+\cos\theta}{\sin\theta}=\frac{2\cos^2\frac\theta2}{2\sin\frac\theta2\cos\frac\theta2}=\cot\frac\theta2\).

\(\therefore u=\mathrm{sech}^{-1}(\sin\theta)=\log\left(\cot\frac\theta2\right)\) (proved).