Question

If $f\left(\theta \right)=\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +\frac{2\pi }{3})$ then $f\left(\frac{\pi }{12}\right)-f^{{}^{\prime }}\left(\frac{\pi }{12}\right)$ equals

• A. $25$
• B. $-15$
• C. $-25$
• D. $15$

Answer 1

The correct option is B $-15$

Here $f\left(\theta \right)=\tan \theta +\tan (\frac{\pi }{3}+\theta )+\tan (\frac{2\pi }{3}+\theta )$

$=\tan \theta +\tan (\frac{\pi }{3}+\theta )-\tan (\frac{\pi }{3}-\theta )$

$=\tan \theta +\frac{\tan \frac{\pi }{3}+\tan \theta }{1-\tan \frac{\pi }{3}\tan \theta }-\frac{\tan \frac{\pi }{3}-\tan \theta }{1+\tan \frac{\pi }{3}\tan \theta }$

$=\tan \theta +\frac{\sqrt{3}+\tan \theta }{1-\sqrt{3}\tan \theta }-\frac{\sqrt{3}-\tan \theta }{1+\sqrt{3}\tan \theta }$

$=\frac{\tan \theta (1-3{\tan }^2\theta )+(\sqrt{3}+\tan \theta )(1+\sqrt{3}\tan \theta )-(\sqrt{3}-\tan \theta )(1-\sqrt{3}\tan \theta )}{1-3{\tan }^2\theta }$

$=3\tan 3\theta$

And $f^{{}^{\prime }}\left(\theta \right)=9{\sec }^{2}3\theta$

Hence

$f\left(\frac{\pi }{12}\right)-f^{{}^{\prime }}\left(\frac{\pi }{12}\right)=3\tan \frac{\pi }{4}-9{\sec }^2\frac{\pi }{4}$

$=3-9\times 2=-15$