$\frac{d}{dx}\left[{\tan }^{-1}(\text{cosec}x+\cot x)\right]=(\text{where }x\in \left(\text{0,2π}\right))$

$\frac{d}{dx}\left[{\tan }^{-1}(\text{cosec}x+\cot x)\right]$

$=\frac{1}{1+{(\text{cosec}x+\cot x)}^2}\times [-\text{cosec}x.\cot x-{\text{cosec}}^{2}x]$

$=\frac{-\text{cosec}x.\cot x-{\text{cosec}}^{2}x}{1+{\text{cosec}}^{2}x+{\cot }^{2}x+2\text{cosec}x.\cot x}$

$=\frac{-\text{cosec}x.[\cot x+\text{cosec}x]}{{\text{2cosec}}^{2}x+2\text{cosec}x.\cot x}$

$=\frac{-\text{cosec}x}{2\text{cosec}x}$

$=-\frac{1}{2}$

(or)

$\frac{d}{dx}{\tan }^{-1}(\frac{1}{\sin x}+\frac{\cos x}{\sin x})$

$=\frac{d}{dx}{\tan }^{-1}\left(\frac{1+\cos x}{\sin x}\right)$

$=\frac{d}{dx}{\tan }^{-1}\left(\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}\right)$

$=\frac{d}{dx}{\tan }^{-1}\left(\tan (\frac{\pi }{2}-\frac{x}{2})\right)$

$=\frac{d}{dx}(\frac{\pi }{2}-\frac{x}{2})=\frac{-1}{2}$