If `L=lim_(xto(pi^(+))/2)(costan^(-1)(tanx))/(x-pi//2)` then `cos(2piL)` is

Answer

To solve the limit \( L = \lim_{x \to \frac{\pi}{2}^+} \frac{\cos(\tan^{-1}(\tan x))}{x - \frac{\pi}{2}} \) and find \( \cos(2\pi L) \), we can follow these steps:

### Step 1: Simplify the expression
We start with the limit:
\[ L = \lim_{x \to \frac{\pi}{2}^+} \frac{\cos(\tan^{-1}(\tan x))}{x - \frac{\pi}{2}} \]
As \( x \) approaches \( \frac{\pi}{2} \) from the right, \( \tan x \) approaches \( -\infty \) because \( \tan x \) is undefined at \( \frac{\pi}{2} \) and approaches negative values from the right. Therefore, \( \tan^{-1}(\tan x) \) will approach \( -\frac{\pi}{4} \) (since \( \tan^{-1}(-1) = -\frac{\pi}{4} \)).

### Step 2: Substitute the limit
Thus, we can rewrite the limit:
\[ L = \lim_{x \to \frac{\pi}{2}^+} \frac{\cos(-\frac{\pi}{4})}{x - \frac{\pi}{2}} \]
Since \( \cos(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \), we have:
\[ L = \lim_{x \to \frac{\pi}{2}^+} \frac{\frac{1}{\sqrt{2}}}{x - \frac{\pi}{2}} \]

### Step 3: Evaluate the limit
As \( x \) approaches \( \frac{\pi}{2}^+ \), the denominator \( x - \frac{\pi}{2} \) approaches \( 0^+ \). Therefore, the limit diverges to \( +\infty \):
\[ L = +\infty \]

### Step 4: Find \( \cos(2\pi L) \)
Since \( L \) is infinite, we can calculate:
\[ \cos(2\pi L) = \cos(2\pi \cdot \infty) \]
The cosine function is periodic with a period of \( 2\pi \), so:
\[ \cos(2\pi L) = \cos(\text{any multiple of } 2\pi) = 1 \]

### Final Answer
Thus, the value of \( \cos(2\pi L) \) is:
\[ \boxed{1} \]