Question

$\frac{\cot x}{\cot x-\cot 3x}+\frac{\tan x}{\tan x-\tan 3x}=$

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$\frac{\cot x}{\cot x-\cot 3x}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{\frac{1}{\tan x}}{\frac{1}{\tan x}-\frac{1}{\tan 3x}}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{\frac{1}{\tan x}}{\frac{\tan 3x-\tan x}{\tan x\tan 3x}}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{1}{\frac{\tan 3x-\tan x}{\tan 3x}}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{\tan 3x}{\tan 3x-\tan x}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{-\tan 3x}{\tan x-\tan 3x}+\frac{\tan x}{\tan x-\tan 3x}$

$=\frac{-\tan 3x+\tan x}{\tan x-\tan 3x}$

$=\frac{\tan x-\tan 3x}{\tan x-\tan 3x}=1$