Question

If tan x + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to

Options

• \[\frac{5\pi}{6}\]

   

• \[\frac{2\pi}{3}\]

   

• \[\frac{\pi}{6}\]

   

• \[\frac{\pi}{3}\]

Answer 1

\[\frac{\pi}{6}\]
We have: 
\[\tan x + \sec x = \sqrt{3} \left[ 0 < x < \pi \right]\]
\[ \Rightarrow sec x + \tan x = \sqrt{3}\]
\[ \Rightarrow \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sqrt{3}\]
\[ \Rightarrow 1 + \sin x=\sqrt{3}\cos x\]
\[\Rightarrow \left( 1 + \sin x \right)^2 = \left( \sqrt{3} \cos x \right)^2 \]
\[ \Rightarrow 1 + \sin^2 x + 2\sin x = 3 \cos^2 x\]
\[ \Rightarrow 1 + \sin^2 x + 2\sin x = 3(1 - \sin^2 x)\]
\[ \Rightarrow 4 \sin^2 x + 2\sin x = 2\]
\[ \Rightarrow 2 \sin^2 x + \sin x - 1 = 0\]
\[ \Rightarrow \sin x = - 1, \frac{1}{2}\]
\[\text{ Since }0 < x < \pi, \sin x \text{ cannot be negative .} \]
\[ \therefore \sin x = \frac{1}{2}\]
\[ \therefore x = \frac{\pi}{6} \]