# If cos(x) = (1/4), where x lies in quadrant 4, how do you find tan(x - pi/4)?

Aug 27, 2016

The reqd. value$= \frac{\sqrt{15} + 1}{\sqrt{15} - 1}$.

#### Explanation:

We have, $\tan \left(x - \frac{\pi}{4}\right) = \frac{\tan x - \tan \left(\frac{\pi}{4}\right)}{1 + \tan x \tan \left(\frac{\pi}{4}\right)}$

$= \frac{\tan x - 1}{1 + \tan x} = \left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$

So, we need $\sin x$ to find the reqd. value.

$\cos x = \frac{1}{4} , x \in {Q}_{I V} \Rightarrow \sin x = - \sqrt{1 - {\cos}^{2} x} = - \frac{\sqrt{15}}{4}$.

Hence, the reqd. value$= \frac{- \frac{\sqrt{15}}{4} - \frac{1}{4}}{- \frac{\sqrt{15}}{4} + \frac{1}{4}} = \frac{\sqrt{15} + 1}{\sqrt{15} - 1}$.