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Q.
If $tan\theta = 3$ and $\theta$ lies in $III$ quadrant, then find the value of $sin\theta$.
Trigonometric Functions
Solution:
Given, $tan\theta=\frac{3}{1}$ and $\theta$ lies in $III$ quadrant.
$\because sec^{2}\theta=1+tan^{2}\theta=10$
$\Rightarrow sec\theta=\pm\sqrt{10}$
Since, $\theta$ lies in III quadrant, so $sec\theta=-\sqrt{10}$
$\Rightarrow cos\theta=\frac{1}{sec\,\theta}=\frac{1}{-\sqrt{10}}$
Now, $sin^{2}\theta=1-cos^{2}\theta=\frac{9}{10}$
$\Rightarrow sin\theta=\pm \sqrt{\frac{9}{10}}$
Since, $\theta$ lies in $III$ quadrant so $sin\theta=-\sqrt{\frac{9}{10}}=\frac{-3}{\sqrt{10}}$