Question

The period of $\frac{\cos \frac{x}{4}+\tan \frac{x}{4}}{1+\tan \frac{x}{2}-\tan x}$

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $8\pi$
• D. $2\pi$

Answer 1

The correct option is C $8\pi$

$f\left(x\right)=\frac{\cos \frac{x}{4}+\tan \frac{x}{4}}{1+\tan \frac{x}{2}-\tan x}=\frac{h\left(x\right)}{g\left(x\right)}$

Period of $f\left(x\right)=\mathrm{LCM}$ of $\left\{h\right(x)$ and $g\left(x\right)\}$

Period of $\cos \left(\frac{x}{4}\right)=8\pi$.

Period of $\tan \left(\frac{x}{4}\right)=4\pi$

$\therefore$ Periad of $\{\cos \left(\frac{x}{4}\right)+\tan \left(\frac{x}{4}\right)\}=8\pi$

Period of $\tan \frac{x}{2}=2\pi$

Period of $\tan x=\pi$

Period of $\{1+\tan \frac{x}{2}-\tan x\}=2\pi$

Period of $f\left(x\right)=8\pi$ ie $\mathrm{LCM}$ of $2\pi$ and $8\pi$

$\therefore$ Answer is $8\pi$