Question

If $f\left(x\right)=\frac{3x-x^3}{1-3x^2},\tan \theta \ne \frac{1}{\sqrt{3}}$ , find the value of $f\left(\tan \theta \right)$ is equal to:

• A. $3\tan \theta$
• B. $\tan 3\theta$
• C. ${\tan }^3\theta$
• D. ${\tan }^4\theta$

Answer 1

Answer: (B)

$\tan 3\theta$

It is given that

$f\left(x\right)=\frac{3x-x^3}{1-3x^2}$

Therefore

$f\left(\tan \theta \right)=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$ where $\tan \theta \ne \frac{1}{\sqrt{3}}$

Hence

$f\left(\tan \theta \right)=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$=\tan 3\theta$.

Hence

$f\left(\tan \theta \right)=\tan 3\theta$.