# How do you evaluate 2((tan^-1(1/3))-tan^-1(-1/7)?

##### 1 Answer
Jul 3, 2016

$\frac{\pi}{4}$.

#### Explanation:

We have to use these Rules : R(1) : tan^-1x+tan^-1y=tan^-1{(x+y)/(1-xy)}; x,y>0, xy<1.

$R \left(2\right) : {\tan}^{-} 1 \left(- x\right) = - \left({\tan}^{-} 1 x\right) , x \in \mathbb{R}$.

The Given Exp. $= 2 {\tan}^{-} 1 \left(\frac{1}{3}\right) - {\tan}^{-} 1 \left(- \frac{1}{7}\right)$
$= {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right) \ldots \ldots \ldots \ldots \ldots . . \left[R \left(2\right)\right]$

$= {\tan}^{-} 1 \left\{\frac{\frac{1}{3} + \frac{1}{3}}{1 - \frac{1}{3} \cdot \frac{1}{3}}\right\} + {\tan}^{-} 1 \left(\frac{1}{7}\right) \ldots \ldots \ldots \ldots \ldots . \left[R \left(1\right)\right]$

$= {\tan}^{-} 1 \left(\frac{2}{3} \cdot \frac{9}{8}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right)$
${\tan}^{-} 1 \left(\frac{3}{4}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right)$
$= {\tan}^{-} 1 \left\{\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}}\right\} \ldots \ldots \ldots \ldots \ldots . \left\{R \left(1\right)\right]$
$= {\tan}^{-} 1 \left\{\frac{\frac{25}{28}}{\frac{25}{28}}\right\}$
$= {\tan}^{-} 1 \left(1\right)$
$= \frac{\pi}{4}$.