# How do you find the exact value of sec^-1(sqrt2)?

Sep 1, 2016

45degrees or $\frac{\pi}{4}$

#### Explanation:

Draw a. Right angled isosceles triangle . The ratio of the sides 1:1:$\sqrt{2}$

Sep 1, 2016

${\sec}^{-} 1 \sqrt{2} = \frac{\pi}{4}$.

#### Explanation:

The Defn. of ${\sec}^{-} 1$ is :

${\sec}^{-} 1 x = \theta , | x | . = 1 \iff \sec \theta = x , \theta \in \left[0 , \pi\right] - \left\{\frac{\pi}{2}\right\}$.

Knowing that, $\sec \left(\frac{\pi}{4}\right) = \sqrt{2} , \frac{\pi}{4} \in \left[0 , 2 \pi\right] - \left\{\frac{\pi}{2}\right\} , {\sec}^{-} 1 \sqrt{2} = \frac{\pi}{4}$.