Question

If $tan\theta =3$ and $\theta$ lies in $III$ quadrant, then find the value of $sin\theta$.

• A. $1/\sqrt{10}$
• B. $2/\sqrt{10}$
• C. $-3/\sqrt{10}$
• D. $-5/\sqrt{10}$

Answer 1

Answer: (C)

Given, $tan\theta =\frac{3}{1}$ and $\theta$ lies in $III$ quadrant.
$\because se{c}^2\theta =1+ta{n}^2\theta =10$
$\Rightarrow sec\theta =\pm \sqrt{10}$
Since, $\theta$ lies in III quadrant, so $sec\theta =-\sqrt{10}$
$\Rightarrow cos\theta =\frac{1}{sec\theta }=\frac{1}{-\sqrt{10}}$
Now, $si{n}^2\theta =1-co{s}^2\theta =\frac{9}{10}$
$\Rightarrow sin\theta =\pm \sqrt{\frac{9}{10}}$
Since, $\theta$ lies in $III$ quadrant so $sin\theta =-\sqrt{\frac{9}{10}}=\frac{-3}{\sqrt{10}}$