# What is the exact value of tan(t+x) given that sin(t)=3/5 and sin(x)=12/13 knowing that t & x are between 0 and pi/2 and that the sides are Pythagorean triples?

## Homework question stumping.. I came up with 56/33 trying to confirm and validate. Thanks!

Jul 27, 2018

$\tan \left(t + x\right) = - \frac{63}{16}$

#### Explanation:

If angle $t$ is $0 < t < \frac{\pi}{2}$, then $t$ is in the first quadrant. Using a right-angled triangle,

$\sin t = \frac{3}{5}$

$\cos t = \frac{4}{5}$ (cos is positive in the first quadrant)

$\tan t = \frac{3}{4}$ (tan is positive in the first quadrant)

If angle $x$ is also $0 < x < \frac{\pi}{2}$, then $x$ is in the first quadrant. Using a right-angled triangle,

$\sin x = \frac{12}{13}$

$\cos x = \frac{5}{13}$ (cos is positive in the first quadrant)

$\tan x = \frac{12}{5}$ (tan is positive in the first quadrant)

$\tan \left(t + x\right)$

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

$= \frac{\frac{3}{4} + \frac{12}{5}}{1 - \frac{3}{4} \times \frac{12}{5}}$

$= \frac{\frac{63}{20}}{1 - \frac{9}{5}}$

$= \frac{63}{20} \times - \frac{5}{4}$

$= - \frac{63}{16}$