If $\sin \theta=\frac{3}{5}$ and $\theta$ is not in the first quadrant, then $15 \sin$ $2 \theta-20 \cos 2…
If $\sin \theta=\frac{3}{5}$ and $\theta$ is not in the first quadrant, then $15 \sin$ $2 \theta-20 \cos 2 \theta-7 \tan 2 \theta=$
-4
-12
12
4
Solution
Given: $\sin \theta=\frac{3}{5}$
Since, $\theta$ is not in first quadrant then $\theta$ will be in second quadrant because $\sin \theta$ is positive in first \& second coordinate.
$\cos \theta=\sqrt{1-\sin ^2 \theta}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$
Now, $\sin 2 \theta=2 \sin \theta \cdot \cos \theta=2 \cdot \frac{3}{5}\left(-\frac{9}{5}\right)=-\frac{24}{25}$
$\begin{aligned} & \cos 2 \theta=2 \cos ^2 \theta-1=2\left(-\frac{4}{5}\right)^2-1=\frac{7}{25} \\ & \tan 2 \theta=-\frac{24}{7}\end{aligned}$
Now, $15 \sin 2 \theta-20 \cos 2 \theta-7 \tan 2 \theta$
$=15 .\left(-\frac{24}{25}\right)-20 \times \frac{7}{25}-7\left(-\frac{24}{7}\right)=4$