# What is the exact value of sec^-1 (sqrt2) and csc^-1 (2)?

Mar 28, 2018

${\sec}^{-} 1 \left(\sqrt{2}\right) = \frac{\pi}{4}$
${\csc}^{-} 1 \left(2\right) = \frac{\pi}{6}$

#### Explanation:

Note:
color(red)((1)sec^-1x=cos^-1(1/x)

color(red)((2)csc^-1x=sin^-1(1/x)

color(red)((3)cos^-1(costheta))=color(red)(theta,where,theta in [0,pi]

color(red)((4)sin^-1(sintheta))=color(red)(theta,where,theta in [-pi/2,pi/2]

Here,

${\sec}^{-} 1 \left(\sqrt{2}\right) = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) \ldots \to A p p l y \left(1\right)$

${\sec}^{-} 1 \left(\sqrt{2}\right) = {\cos}^{-} 1 \left(\cos \left(\frac{\pi}{4}\right)\right)$

${\sec}^{-} 1 \left(\sqrt{2}\right) = \frac{\pi}{4} \in \left[0 , \pi\right] \ldots \ldots \ldots \ldots . \to A p p l y \left(3\right)$

Now,

${\csc}^{-} 1 \left(2\right) = {\sin}^{-} 1 \left(\frac{1}{2}\right) \ldots . . \to A p p l y \left(2\right)$

${\csc}^{-} 1 \left(2\right) = {\sin}^{-} 1 \left(\sin \left(\frac{\pi}{6}\right)\right)$

${\csc}^{-} 1 \left(2\right) = \frac{\pi}{6} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \ldots \ldots \ldots . \to A p p l y \left(4\right)$