Question

$\frac{d}{dx}\left(\frac{1+\cot x}{1-\cot x}\right)=$

Answer 1

$\frac{d}{dx}\left(\frac{1+\cot x}{1-\cot x}\right)=\frac{\frac{d}{dx}(1+\cot x)-\frac{d}{dx}(1-\cot x)}{{(1-\cot x)}^2}=\frac{-{\csc }^{2}x-\left({\csc }^{2}x\right)}{{(1-\cot x)}^2}=\frac{-2{\csc }^{2}x}{1+{\cot }^{2}x-2\cot x}=\frac{-2{\csc }^{2}x}{{\csc }^{2}x-2\cot x}=\frac{\frac{-2}{{\sin }^{2}x}}{\frac{1}{{\sin }^{2}x}-\frac{2\cos x}{\sin x}}=\frac{-2}{1-2\cos x.\sin x}=\frac{-2}{1-\sin 2x}=\frac{2}{\sin 2x-1}$