Question

If $L=\underset{x\to \frac{\pi }{2^{+}}}{lim}\frac{\cos \left({\tan }^{-1}\left(\tan x\right)\right)}{x-\frac{\pi }{2}}$ then $\cos \left(2\pi L\right)$ is:

Answer 1

$L=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{\cos \left({\tan }^{-1}\left(\tan \left(x\right)\right)\right)}{x-\frac{\pi }{2}}$ (of the form $\frac{0}{0}$)

$L=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}\frac{-\sin \left({\tan }^{-1}\left(\tan x\right)\right).\left(\frac{1}{1+{\tan }^{2}x}\right).{\sec }^{2}x}{1}$ [L Hospital Rule)

$L=1(\because x\to 0\Rightarrow \tan x\to -\infty )$

Then, $\cos 2\pi L=\cos \left(2\pi \right)=1$