# How do you find tan (x+y) if tan x=5/4 and sec y=2?

Jul 13, 2016

$\tan \left(x + y\right) = \frac{5 + 4 \sqrt{3}}{4 - 5 \sqrt{3}} ,$ or,$= \frac{5 - 4 \sqrt{3}}{4 + 5 \sqrt{3}}$

#### Explanation:

Given that $\sec y = 2$, we use the Identity $: {\sec}^{2} y = 1 + {\tan}^{2} y$ to get, $\tan y = \pm \sqrt{{\sec}^{2} y - 1} = \pm \sqrt{4 - 1} = \pm \sqrt{3.}$

Now, $\tan \left(x + y\right) = \frac{\tan x + \tan y}{1 - \tan x \cdot \tan y} = \frac{\frac{5}{4} \pm \sqrt{3}}{1 - \frac{5}{4} \cdot \left(\pm \sqrt{3}\right)}$

=(5+-4sqrt3)/(4-(+-5sqrt3)

Thus, $\tan \left(x + y\right) = \frac{5 + 4 \sqrt{3}}{4 - 5 \sqrt{3}} ,$ or,$= \frac{5 - 4 \sqrt{3}}{4 + 5 \sqrt{3}}$