# Find exact value of tan(t+x) if sin(t)=3/5 and sin(x)=12/13 and t and x are between 0 pi/2?

## between 0 and pi/2 Help much appreciated!

May 19, 2018

$- 3.94$ or $- \frac{63}{16}$

#### Explanation:

So, by formula,
$\tan \left(t + x\right) = \frac{\tan t + \tan x}{1 - \tan t \cdot \tan x}$

Proof
$\tan \left(t + x\right) = \sin \frac{t + x}{\cos} \left(t + x\right)$

$= \frac{\sin t \cdot \cos x + \cos t \cdot \sin x}{\cos t \cdot \cos x - \sin t \cdot \sin x}$

divide Numerator and Denominator by $\cos t \cdot \cos x$ and separate the terms, and you"ll get,
$\tan \left(t + x\right) = \frac{\sin \frac{t}{\cos} t + \sin \frac{x}{\cos}}{1 - \sin \frac{t}{\cos} t \cdot \sin \frac{x}{\cos} x}$

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

Coming back to the problem,
as you know $\sin t = \frac{3}{5}$,
$\tan t = \sin \frac{t}{\cos} t$

$= \sin \frac{t}{\sqrt{1 - {\sin}^{2} t}}$

$= \frac{3}{5} / \frac{4}{5} = \frac{3}{4}$

Similarly,
$\tan x = \sin \frac{x}{\cos} x$

$= \sin \frac{x}{\sqrt{1 - {\sin}^{2} x}}$

$= \frac{12}{13} / \frac{5}{13} = \frac{12}{5}$

From the above 2,
$\tan \left(t + x\right) = \frac{\frac{3}{4} + \frac{12}{5}}{1 - \frac{3}{4} \cdot \frac{12}{5}}$

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

$= - 3.94$

Hope this Helps! :)