Question

If $\tan \theta =3$ and $\theta$ lies in third quadrant, then the value of $\sin \theta$ is

• A. $\frac{3}{\sqrt{10}}$
• B. $\frac{-1}{\sqrt{10}}$
• C. $\frac{1}{\sqrt{10}}$
• D. $\frac{-3}{\sqrt{10}}$

Answer 1

The correct option is D $\frac{-3}{\sqrt{10}}$

Apply trigonometric identity of $\tan \theta$ & $\sec \theta$.

Given expression : $\tan \theta =3$

$\Rightarrow {\tan }^2\theta =9$

$\Rightarrow {\sec }^2\theta -1=9$

$(\because {\tan }^2\theta ={\sec }^2\theta -1)$

$\Rightarrow {\sec }^2\theta =10$

$\Rightarrow {\cos }^2\theta =\frac{1}{10}$ …(1)

$(\because \sec \theta =\frac{1}{\cos \theta })$

Apply trigonometric identity related to $\sin \theta$ & $\cos \theta$ and use the obtained value to evaluate the unknown

$\because {\sin }^2\theta =1-{\cos }^2\theta$

$\Rightarrow {\sin }^2\theta =1-\frac{1}{10}=\frac{9}{10}$ (From (1))

$\Rightarrow \sin \theta =\pm \frac{3}{\sqrt{10}}$

As $\theta$ lies in third quadrant, so $\sin \theta$ is negative.

$\Rightarrow \sin \theta =-\frac{3}{\sqrt{10}}$

Hence, option (C) is correct.