Question

If ${90}^o<\theta <{180}^o,\tan \theta =\frac{-8}{15}$ then $\sin 2\theta$

Answer 1

$\tan \theta =\frac{-8}{15}$

$\sin \theta =\frac{-8}{\sqrt{226+64}}=\frac{8}{17}$

$\cos \theta =\frac{-15}{17}$ ($\cos \theta$ will be negative as $\theta$ is in second quadrant).

Now, $\sin 2\theta =2\sin \theta \cos \theta =2\times \frac{8}{17}\times \frac{-15}{17}=\frac{249}{289}$