Question

For $x\in (0,\frac{3}{2}),$ let $f\left(x\right)=\sqrt{x},g\left(x\right)=\tan x$ and $h\left(x\right)=\frac{1-x^2}{1+x^2}.$ If $\varphi \left(x\right)=\left(\right(hof\left)og\right)\left(x\right),$ then $\varphi \left(\frac{\pi }{3}\right)$ is equal to

• A. $\tan \left(\frac{\pi }{12}\right)$
• B. $\tan \left(\frac{5\pi }{12}\right)$
• C. $\tan \left(\frac{7\pi }{12}\right)$
• D. $\tan \left(\frac{11\pi }{12}\right)$

Answer 1

The correct option is D $\tan \left(\frac{11\pi }{12}\right)$

$f\left(x\right)=\sqrt{x}\dots \left(I\right)$
$g\left(x\right)=\tan x\dots \left(II\right)$
$h\left(x\right)=\frac{1-x^2}{1+x^2}\dots \left(III\right)$
for $x\in (0,\frac{3}{2})$
Also,
$\varphi \left(x\right)=\left(\right(hof\left)og\right)\left(x\right)=h\left(f\right(g\left(x\right)\left)\right)$
$=h\left(f\right(\tan x\left)\right)=h\left(\sqrt{\tan x}\right)$
$=\frac{1-(\sqrt{\tan x}{)}^2}{1+(\sqrt{\tan x}{)}^2}=\frac{1-\tan x}{1+\tan x}$

Or, $\varphi \left(x\right)=\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan x\tan \frac{\pi }{4}}$

$\Rightarrow \varphi \left(x\right)=\tan (\frac{\pi }{4}-x)$ $[\because \frac{\tan A-\tan B}{1+\tan A\tan B}=\tan (A-B\left)\right]$

Hence,
$\varphi \left(\frac{\pi }{3}\right)=\tan (\frac{\pi }{4}-\frac{\pi }{3})$
$=\tan (-\frac{\pi }{12})=\tan (\pi -\frac{\pi }{12})\therefore \varphi \left(\frac{\pi }{3}\right)=\tan \left(\frac{11\pi }{12}\right)$