JEE Main 2019 (Online) 12th April Morning ShiftQuestion 1
For $x\in\left(0,\frac32\right)$, let $f(x)=\sqrt x$, $g(x)=\tan x$ and $h(x)=\frac{1-x^2}{1+x^2}$. If $\phi(x)=(h\circ f)\circ g(x)$, then $\phi\left(\frac\pi3\right)$ is equal to:
Functions
Solution
$\phi(x)=\frac{1-\tan x}{1+\tan x}=\tan\left(\frac\pi4-x\right)$. At $x=\frac\pi3$: $\tan\left(-\frac\pi{12}\right)=\tan\frac{11\pi}{12}$.