Question

If $f:(-\frac{\pi }{2},\frac{\pi }{2})\to (-\infty ,\infty )$ is defined by $f\left(x\right)=\tan x$ , then $f^{-1}(2+\sqrt{3})=$

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{4}$
• C. $\frac{5\pi }{12}$
• D. $\frac{\pi }{8}$

Answer 1

Answer: (C)

$\frac{5\pi }{12}$

$f^{-1}\left(x\right)={\tan }^{-1}x$

$f^{-1}(2+\sqrt{3})$

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

we know that $\tan \frac{\pi }{4}=1$ and using half angle formula considering $\frac{\pi }{4}$ we get $\tan \frac{\pi }{8}=\frac{2-\sqrt{2}}{\sqrt{2}}$

Considering for $\frac{5\pi }{12}$

$\tan \left(\frac{5\pi }{12}\right)=\tan (\frac{6\pi }{12}-\frac{\pi }{12})=\tan (\frac{\pi }{2}-\frac{\pi }{12})=\cot \left(\frac{\pi }{12}\right)=\frac{1}{\tan \frac{\pi }{12}}$

Let $\frac{\pi }{12}=t$

$\tan 2t=\frac{\left(2\tan t\right)}{1-{\tan }^{2}t}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$\Rightarrow {\tan }^{2}t+2\sqrt{3}\tan t-1=0$

$D=b^2-4ac=12+4\Rightarrow b=\pm 4$

Solving through quadratic formula

$\therefore \tan t=-\sqrt{3}\pm 2$

$\because \tan \left(\frac{\pi }{12}\right)$ lies in first quadrant only positive value would be taken

$\therefore \tan \frac{5\pi }{12}=\frac{1}{\tan \frac{\pi }{12}}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$

From (i)

$\therefore f^{-1}(2+\sqrt{3})$

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

$=n\pi +\frac{5\pi }{12}\forall nϵz$...................$\because \tan \frac{5\pi }{12}=2+\sqrt{3}$

But as $x\in [\frac{-\pi }{2},\frac{\pi }{2}]\Rightarrow f^{-1}(2+\sqrt{3})=\frac{5\pi }{12}$