Question

Prove the following identities (1-16)

$\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=\left(\sec x\mathrm{cossec}x+1\right)$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x} \\ =\frac{{\displaystyle \frac{\sin x}{\cos x}}}{1-{\displaystyle \frac{\cos x}{\sin x}}}+\frac{{\displaystyle \frac{\cos x}{\sin x}}}{1-{\displaystyle \frac{\sin x}{\cos x}}} \\ =\frac{{\displaystyle \frac{\sin x}{\cos x}}}{{\displaystyle \frac{\sin x-\cos x}{\sin x}}}+\frac{{\displaystyle \frac{\cos x}{\sin x}}}{{\displaystyle \frac{\cos x-\sin x}{\cos x}}} \\ =\frac{\sin x}{\cos x}\times \frac{\sin x}{\sin x-\cos x}+\frac{\cos x}{\sin x}\times \frac{\cos x}{\cos x-\sin x} \\ =\frac{\sin x}{\cos x}\times \frac{\sin x}{\sin x-\cos x}+\frac{\cos x}{\sin x}\times \frac{\cos x}{-\left(\sin x-\cos x\right)} \\ =\frac{{\sin }^{2}x}{\cos x\left(\sin x-\cos x\right)}-\frac{{\cos }^{2}x}{\sin x\left(\sin x-\cos x\right)} \\ =\frac{{\sin }^{3}x-{\cos }^{3}x}{\sin x\cos x\left(\sin x-\cos x\right)} \\ =\frac{\left(\sin x-\cos x\right)\left[{\sin }^{2}x+{\cos }^{2}x+\sin x\cos x\right]}{\sin x\cos x\left(\sin x-\cos x\right)} \\ =\frac{1\times \left[1+\sin x\cos x\right]}{\sin x\cos x} \\ =\frac{1+\sin x\cos x}{\sin x\cos x} \\ =\frac{1}{\sin x\cos x}+\frac{\sin x\cos x}{\sin x\cos x} \\ =\frac{1}{\sin x}\times \frac{1}{\cos x}+1 \\ =\mathrm{cosec}x\times \sec x+1 \\ =\left(\sec x\mathrm{co}\sec x+1\right) \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$