Question

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

• A. $\frac{5\pi }{6}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is C

$\frac{\pi }{6}$

$\frac{\pi }{6}$

We have:

$tan\theta +sec\theta =\sqrt{3}[0<\theta <\pi ]$

$\Rightarrow sec\theta +tan\theta =\sqrt{3}$

$\Rightarrow \frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }=\sqrt{3}$

$\Rightarrow 1+sin\theta =\sqrt{3}cos\theta$

$\Rightarrow (1+sin\theta {)}^2=(\sqrt{3}cos\theta {)}^2$

$\Rightarrow 1+si{n}^2\theta +2sin\theta =3co{s}^2\theta$

$\Rightarrow 1+si{n}^2\theta +2sin\theta =3(1-si{n}^2\theta )$

$\Rightarrow 4si{n}^2\theta +2sin\theta =2$

$\Rightarrow 2si{n}^2\theta +sin\theta -1=0$

$\Rightarrow sin\theta =-1,\frac{1}{2}$

Since $0<\theta <\pi ,sin\theta$ cannot be negative.

$\therefore sin\theta =\frac{1}{2}$

$\therefore \theta =\frac{\pi }{6}$