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 $\varphi \left(x\right)\left(\frac{\pi }{3}\right)=\left(\left(hof\right)og\right)\left(x\right)$. If $\varphi$ is equal to tan $\frac{p\pi }{q}$ then $p+q$ is

Answer 1

Answer: 23

$\because \varphi \left(x\right)=\left(\left(hof\right)og\right)\left(x\right)$
$\because \varphi \left(\frac{\pi }{3}\right)=h\left(f\left(g\left(\frac{\pi }{3}\right)\right)\right)=h\left(f\left(\sqrt{3}\right)\right)=h\left(3^{1/4}\right)$
$=\frac{1-\sqrt{3}}{1+\sqrt{3}}=\frac{1}{2}\left(1+3-2\sqrt{3}\right)=\sqrt{3}-2=-\left(-\sqrt{3}+2\right)$
$=-tan15^{\circ }=tan\left(180^{\circ }-15^{\circ }\right)=tan\left(\pi -\frac{\pi }{12}\right)=tan\frac{11\pi }{12}$
Now, tan $\frac{p\pi }{q}=tan\frac{11\pi }{12}$
$\Rightarrow p=11,q=12$
Hence, $p+q=11+12=23$