Question

Prove:

$\frac{\tan Q}{1-\cot Q}+\frac{\cot Q}{1+-\cot Q}=1+\tan Q+\cot Q$

Answer 1

$ L H S =

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

$\frac{\cot x}{1-\tan x}+\frac{\tan x(-\tan x)}{(1-\cot x)(-\tan x)}$

$\frac{\cot x}{1-\tan x}-\frac{{\tan }^{2}x}{1-\tan x}$

$\frac{\cot x-{\tan }^{2}x}{1-\tan x}=\frac{1-{\tan }^{3}x}{\tan x(1-\tan x)}$

$\frac{(1-\tan x)(1+\tan x+{\tan }^{2}x)}{(1-\tan x)\left(\tan x\right)}$

$1+\cot x+\tan x=$ R H S

Hence proved.