Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

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

Solution:

$\mathrm{LHS}=\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}$

$=\frac{\frac{\sin x}{\cos x}}{1-\frac{\cos x}{\sin x}}+\frac{\frac{\cos x}{\sin x}}{1-\frac{\sin x}{\cos x}}$

$=\frac{\frac{\sin x}{\cos x}}{\frac{\sin x-\cos x}{\sin x}}+\frac{\frac{\cos x}{\sin x}}{\frac{\cos x-\sin x}{\cos x}}$

$=\frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x-\cos x}+\frac{\cos x}{\sin x} \times \frac{\cos x}{\cos x-\sin x}$

$=\frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x-\cos x}+\frac{\cos x}{\sin x} \times \frac{\cos x}{-(\sin x-\cos x)}$

$=\frac{\sin ^{2} x}{\cos x(\sin x-\cos x)}-\frac{\cos ^{2} x}{\sin x(\sin x-\cos x)}$

$=\frac{\sin ^{3} x-\cos ^{3} x}{\sin x \cos x(\sin x-\cos x)}$

$=\frac{(\sin x-\cos x)\left[\sin ^{2} x+\cos ^{2} x+\sin x \cos x\right]}{\sin x \cos x(\sin x-\cos x)}$

$=\frac{1 \times[1+\sin x \cos x]}{\sin x \cos x}$

$=\frac{1+\sin x \cos x}{\sin x \cos x}$

$=\frac{1}{\sin x \cos x}+\frac{\sin x \cos x}{\sin x \cos x}$

$=\frac{1}{\sin x} \times \frac{1}{\cos x}+1$

$=\operatorname{cosec} x \times \sec x+1$

$=(\sec x \operatorname{cosec} x+1)$

$=$ RHS

Hence proved.

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