# How do you find the exact value of tan [ arctan (1/6) + arccos (3/5) ]?

Jun 21, 2016

#27/14 and -21/22.

#### Explanation:

Let $a = a r c \tan \left(\frac{1}{6}\right)$. Then, $\tan a = \frac{1}{6} > 0$.

a is in either first quadrant or in the third.

Let $b = a r c \cos \left(\frac{3}{5}\right)$. Then, $\cos b = \frac{3}{5} > 0$.

b is in either first quadrant or in the fourth..

Accordingly, $\sin a = . \pm \frac{4}{5} \mathmr{and} \tan a = \pm \frac{4}{3}$.

Now, the given expression is

$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

$= \frac{\left(\frac{1}{6}\right) \pm \left(\frac{4}{3}\right)}{1 - \left(\pm\right) \left(\frac{1}{6}\right) \left(\frac{4}{3}\right)}$

$= \frac{27}{14} \mathmr{and} - \frac{21}{22}$.

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