# How do you evaluate sec^-1(sec((5pi)/6))?

Jul 14, 2016

Answer $= 5 \frac{\pi}{6}$

#### Explanation:

$\sec 5 \frac{\pi}{6} = \sec \left(\pi - \frac{\pi}{6}\right) = - \sec \left(\frac{\pi}{6}\right) = - \frac{2}{\sqrt{3}}$

Now, ${\sec}^{-} 1 \left(- x\right) = \pi - {\sec}^{-} 1 x , | x | \ge 1$

Hence, ${\sec}^{-} 1 \left(\sec 5 \frac{\pi}{6}\right) = {\sec}^{-} 1 \left(- \frac{2}{\sqrt{3}}\right) = \pi - {\sec}^{-} 1 \left(\frac{2}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = 5 \frac{\pi}{6.}$

Otherwise

If a fun. $f : A \rightarrow B$ is an injection, i.e., one-to-one, then

$f \left(x\right) = f \left(y\right) \Rightarrow x = y , \forall x , y \in A .$

So, if ${\sec}^{-} 1 \left(\sec 5 \frac{\pi}{6}\right) = \theta , t h e n , \sec \theta = \sec 5 \frac{\pi}{6.} \ldots . . \left(i\right)$

Now, $\sec$ is $1 - 1$ in $\left[0 , \pi\right] - \left\{\frac{\pi}{2}\right\}$, &, $5 \frac{\pi}{6} \in \left[0 , \pi\right] - \left\{\frac{\pi}{2}\right\}$, from $\left(i\right)$ we conclude that $\theta \ldots . \left[= {\sec}^{-} 1 \left(\sec 5 \frac{\pi}{6}\right)\right] \ldots = 5 \frac{\pi}{6}$