Question

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

Answer 1

$\because \phi \left(x\right)= \left(\left(hof\right)og\right)\left(x\right)$
$\because\phi \left(\frac{\pi}{3}\right) =h\left(f\left(g\left(\frac{\pi}{3}\right)\right)\right)=h\left(f\left(\sqrt{3}\right)\right) =h\left(3^{1/4}\right)$
$= \frac{1-\sqrt{3}}{1+\sqrt{3}} = \frac{1}{2} \left(1 +3-2\sqrt{3}\right) =\sqrt{3}-2 = -\left(-\sqrt{3}+2\right)$
$ = -tan 15^{\circ} =tan \left(180^{\circ} -15^{\circ}\right) = tan \left(\pi-\frac{\pi}{12}\right) =tan \frac{11\pi}{12}$
Now, tan $\frac{p\pi}{q} = tan \frac{11\pi}{12}$
$\Rightarrow p= 11, q=12$
Hence, $p +q =11 +12 =23 $