Question

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

Answer 1

$\tan x=\frac{\sin x}{\cos x}$

$\cot x=\frac{\cos x}{\sin x}$

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

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

$\Rightarrow \frac{{\sin }^{2}x}{(\sin x-\cos x)\cos x}-\frac{{\cos }^{2}x}{\left(\sin x\right)(\sin x-\cos x)}$

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

$a^3-b^3=(a-b)(a^2{b}^2+ab)$

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

$\because {\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow \frac{1+\sin x\cos x}{\sin x\cos x}⟶\left(1\right)$

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

$\Rightarrow \frac{\sin x\cos x+{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}$

$\Rightarrow \frac{1+\sin x\cos x}{\sin x\cos x}⟶\left(2\right)$

LHS=RHS

Hence, proved.