Question

If $\tan \theta =\sqrt{3}$ and 'θ' lies in the third quadrant, then $\sin \theta +\cos \theta$ is

• A.
  $\frac{1-\sqrt{3}}{2}$
• B.
  $\frac{-1-\sqrt{3}}{2}$
• C.
  $\frac{1+\sqrt{3}}{2}$
• D.
  $\frac{-1+\sqrt{3}}{2}$

Answer 1

The correct option is B

$\frac{-1-\sqrt{3}}{2}$
Here, the angle θ lies in the third quadrant. Thus, both sine and cosine of the angle will be negative.
Now, $\tan \theta =\sqrt{3}\mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{the}\mathrm{identity}{\sec }^2\theta =1+{\tan }^2\theta \Rightarrow {\sec }^2\theta =1+3\Rightarrow {\sec }^2\theta =4\Rightarrow \sec \theta =\pm 2\mathrm{But},\mathrm{since}\theta \mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}\sec \theta =-2\Rightarrow \cos \theta =-\frac{1}{2}\mathrm{and}\sin \theta =-\frac{\sqrt{3}}{2}\mathrm{thus},\sin \theta +\cos \theta =\frac{-1-\sqrt{3}}{2}$
Thus, Option d. is correct.