Trigonometry! #25

$\tan\theta=-\frac{4}{3}$

$\cot^{2}\theta=\frac{9}{16}$

$\csc^{2}\theta-1=\frac{9}{16}$

$\frac{1}{\sin^{2}\theta}=\frac{9}{16}+1$

$\sin^{2}\theta=\frac{16}{9+16}$

$\sin\theta=\sqrt{\frac{16}{25}}$

$\boxed{\sin\theta=\pm\frac{4}{5}}$

$\tan \theta = -\frac{4}{3} = \frac{\sin \theta}{\cos \theta} \quad\Leftrightarrow\quad \cos \theta = -\frac{3}{4} \sin \theta$

$1 = \sin^2 \theta + \cos^2 \theta = \sin^2 \theta \cdot \left( 1 + \frac{9}{16} \right) = \sin^2 \theta \cdot \frac{25}{16} \quad\Leftrightarrow\quad \sin^2 \theta = \frac{16}{25} \;\Leftrightarrow\; \sin \theta = \pm \frac{4}{5}$