Question

Find the principle value of:(a)${sec}^{-1}\left(\frac{2}{\sqrt{3}}\right)$ (b)$se{c}^{-1}\left(-2\right).$

Answer 1

We have,

Part (1):-

${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Let

${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=A$

$\Rightarrow \sec A=\frac{2}{\sqrt{3}}$

$\Rightarrow \sec A=\sec \frac{\pi }{6}$

$\Rightarrow A=\frac{\pi }{6}$

Or

${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=A$

$\Rightarrow \sec A=\frac{2}{\sqrt{3}}$

$\Rightarrow \sec A=\sec (2\pi -\frac{\pi }{6})\therefore \sec (2\pi -\theta )=\sec \theta$

$\Rightarrow A=\frac{11\pi }{6}$

Hence, the principal value of ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$is $\frac{\pi }{6},\frac{11\pi }{6}$

Hence, this is the answer.

Part (2):-

${\sec }^{-1}(-2)$

Let

${\sec }^{-1}(-2)=B$

$\sec B=-2$

$\sec B=-\sec \frac{\pi }{3}$

$\sec B=\sec (\pi +\frac{\pi }{3})(\therefore \sec (\pi +\theta )=-\sec \theta )$

$\sec B=\sec \left(\frac{4\pi }{3}\right)$

$B=\frac{4\pi }{3}$

Or

${\sec }^{-1}(-2)=B$

$\sec B=-2$

$\sec B=-\sec \frac{\pi }{3}$

$\sec B=\sec (\pi -\frac{\pi }{3})(\therefore \sec (\pi -\theta )=-\sec \theta )$

$\sec B=\sec \left(\frac{2\pi }{3}\right)$

$B=\frac{2\pi }{3}$

Hence, the principal value of ${\sec }^{-1}\left(2\right)$is $\frac{\pi }{3},\frac{2\pi }{3}$

Hence, this is the answer.