Question

If $\tan \theta =4$, then $\left(\frac{\tan \theta }{\frac{{\sin }^3\theta }{\cos \theta }+\sin \theta \cos \theta }\right)$ is equal to:

• A. $0$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $1$

Answer 1

The correct option is D $1$

$\tan \theta =\frac{4}{1}=\frac{p}{B}$
$H=\sqrt{P^2+B^2}=\sqrt{16+1}=\sqrt{17}$
So, $\sin \theta =\frac{P}{H}=\frac{4}{\sqrt{17}}$, $\cos \theta =\frac{B}{H}=\frac{1}{\sqrt{17}}$
$\therefore$ $\frac{\tan \theta }{\frac{{\sin }^3\theta }{\cos \theta }+\sin \theta \cos \theta }$
$=\frac{4}{\frac{{\left(\frac{4}{\sqrt{17}}\right)}^3}{\frac{1}{\sqrt{17}}}+\frac{4}{\sqrt{17}}\times \frac{1}{\sqrt{17}}}$
$=\frac{4}{\frac{64}{1\sqrt{17}7}\times \sqrt{17}+\frac{4}{17}}=\frac{4}{\frac{64+4}{17}}=\frac{68}{68}=1$