Question

If $\tan \theta=-\frac{4}{3},$ then $\sin \theta$ is :

• A. $-\frac{4}{5}$ but not $\frac{4}{5}$
• B. $-\frac{4}{5}$ or $\frac{4}{5}$
• C. $\frac{4}{5}$ but not $-\frac{4}{5}$
• D. none of these

Answer 1

Answer: (B)

Key Idea : $\tan \theta$ is negative in second and fourth quadrants.
$\tan 0=\cdots \frac{4}{3}$
Here $p=4$ and $b=3$
$\therefore h=\sqrt{p^{2}+b^{2}}=\sqrt{16+9}=\sqrt{25}$
$\Rightarrow h=5$
$\therefore \sin \theta=\frac{p}{h}=\frac{4}{5}$
But $\tan \theta$ is negative which is possible only if $\theta$ lie in second and fourth quadrants.
$\therefore \sin \theta$ may be $\frac{4}{5}$ or $-\frac{4}{5}$.