If + =3,0<x< , then x is equal to: - Vidyadip

MCQ

If $\tan\text{x}+\sec\text{x}=\sqrt{3},0<\text{x}<\pi,$ then $x$ is equal to:

A
$\frac{5\pi}{6}$
B
$\frac{2\pi}{3}$
✓
$\frac{\pi}{6}$
D
$\frac{\pi}{3}$

✓

Answer

Correct option: C.

$\frac{\pi}{6}$

We have:
$\tan\text{x}+\sec\text{x}=\sqrt{3}$ $[0,<\text{x}<\pi]$
$\Rightarrow\sec\text{x}+\tan\text{x}=\sqrt{3}$
$\Rightarrow\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}=\sqrt{3}$
$\Rightarrow1+\sin\text{x}=\sqrt{3}\cos\text{x}$
$\Rightarrow(1+\sin\text{x})^2=(\sqrt{3}\cos\text{x})^2$
$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3\cos^2\text{x}$
$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3(1-\sin^2\text{x})$
$\Rightarrow4\sin^2\text{x}+2\sin\text{x}=2$
$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=-1,\frac{1}{2}$
since $0<\text{x}<\pi,\ \sin\text{x}$ cannot be negative.
$\therefore\sin\text{x}=\frac{1}{2}$
$\therefore\text{x}=\frac{\pi}{6}$

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