Question

How do I find the value of $sec\left(\frac{5\pi }{6}\right)$?

Answer 1

Finding the value of $\sec \left(\frac{5\mathrm{\pi }}{6}\right)$:

Given, $\sec \left(\frac{5\mathrm{\pi }}{6}\right)$

$$\begin{gathered} \therefore \sec \left(\frac{5\pi }{6}\right)=\frac{1}{\cos \left({\displaystyle \frac{5\mathrm{\pi }}{6}}\right)}\left[\because \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\right] \\ =\frac{1}{\cos \left(\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{6}}\right)} \\ =\frac{1}{-\cos \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}\left[\because \cos \left(\pi -\theta \right)=-\cos \left(\theta \right)\right] \\ =-\frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}}\left[\because \cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\right] \\ =-\frac{2}{\sqrt{3}} \end{gathered}$$

$=-\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$ [Multiplying the numerator and the denominator by $\sqrt{3}$]

$=-\frac{2\sqrt{3}}{3}$

Hence, the value of $\sec \left(\frac{5\mathrm{\pi }}{6}\right)$ is $-\frac{2\sqrt{3}}{3}$.