# How do you calculate sin(tan^-1(3/4))?

Jul 11, 2016

$\frac{3}{5.}$

#### Explanation:

If we write ${\tan}^{-} 1 \left(\frac{3}{4}\right) = \theta$, then, by defn. of ${\tan}^{-} 1$ fun., we get, $\tan \theta = \frac{3}{4} , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

Since, $\tan \theta > 0 , \theta \notin \left(- \frac{\pi}{2} , 0\right)$, but, $\theta \in \left(0 , \frac{\pi}{2}\right)$

Now, desired value $= \sin \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = \sin \theta ,$ where, $\tan \theta = \frac{3}{4.}$

Now, $\tan \theta = \frac{3}{4} \Rightarrow \cot \theta = \frac{4}{3} \Rightarrow {\csc}^{2} \theta = 1 + {\cot}^{2} \theta = 1 + {\left(\frac{4}{3}\right)}^{2} = \frac{25}{9} \Rightarrow \csc \theta = \pm \frac{5}{3} \Rightarrow \sin \theta = \pm \frac{3}{5.}$

As, $\theta \in \left(o , \frac{\pi}{2}\right) , \sin \theta > 0 , s o , \sin \theta = + \frac{3}{5}$

The reqd. value $= \frac{3}{5.}$