cot[pi/4 - 2cot-1 3] =7 - Maths - Inverse Trigonometric Functions

Question

cot[pi/4 - 2cot-1 3] =7

Manmeet Singh · Accepted Answer

To prove:

$\cot (\frac{π}{4} - 2 \cot^{- 1} 3) = 7$

Take LHS:

$\cot (\frac{π}{4} - 2 \cot^{- 1} 3) = \tan (\frac{π}{2} - (\frac{π}{4} - 2 \cot^{- 1} 3)) (\sin c e \cot x = \tan (\frac{π}{2} - x)) = \tan (\frac{π}{4} + 2 \cot^{- 1} 3) U \sin g \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \cdot \tan y} = \frac{\tan (\frac{π}{4}) + \tan (2 \cot^{- 1} 3)}{1 - \tan (\frac{π}{4}) \cdot \tan (2 \cot^{- 1} 3)} = \frac{1 + \tan (2 \cot^{- 1} 3)}{1 - \tan (2 \cot^{- 1} 3)} ... (1) S i n c e, \tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x} \tan (2 \cot^{- 1} 3) = \frac{2 \tan (\cot^{- 1} 3)}{1 - \tan^{2} (\cot^{- 1} 3)} S i n c e, \cot^{- 1} x = \tan^{- 1} (\frac{1}{x}) = \frac{2 \tan (\tan^{- 1} (\frac{1}{3}))}{1 - \tan^{2} (\tan^{- 1} (\frac{1}{3}))} = \frac{\frac{2}{3}}{1 - (\frac{1}{3})^{2}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{3}{4} P u t t h i s i n (1), = \frac{1 + \frac{3}{4}}{1 - \frac{3}{4}} = \frac{\frac{7}{4}}{\frac{1}{4}} = 7 = R H S$