# How do you evaluate cos[arc tan(-2/3)]?

Aug 14, 2016

$\frac{3}{\sqrt{13}}$

#### Explanation:

Let #a = arc tan (-2/3) in Q4, by the convention of choosing the

principal value, when the tangent is negative. In Q4, cosine is

positive,

Now, the given expression is

$\cos a = \frac{3}{\sqrt{{2}^{2} + {3}^{2}}} = \frac{3}{\sqrt{13}}$.

Had we chosen $a \in Q 4$, from the general value ( wherein cosine is

negative ), the answer would become $- \frac{3}{\sqrt{13}}$.