The value of `lim_(xto (pi)/2) sqrt((tanx-sin{tan^(-1)(tanx)})/(tanx+cos^(2)(tanx)))` is ………..

Answer

To solve the limit problem step by step, we start with the expression given in the question:

\[ \lim_{x \to \frac{\pi}{2}} \sqrt{\frac{\tan x - \sin(\tan^{-1}(\tan x))}{\tan x + \cos^2(\tan x)}} \]

### Step 1: Analyze the limit as \( x \to \frac{\pi}{2} \)

As \( x \) approaches \( \frac{\pi}{2} \):

- \(\tan x\) approaches infinity.
- \(\sin(\tan^{-1}(\tan x))\) approaches \(\sin(\frac{\pi}{2}) = 1\).
- \(\cos^2(\tan x)\) approaches \(\cos^2(\infty)\), which oscillates between 0 and 1.

Thus, we have:

\[ \tan x - \sin(\tan^{-1}(\tan x)) \to \infty - 1 = \infty \]
\[ \tan x + \cos^2(\tan x) \to \infty + \text{finite} = \infty \]

This gives us the indeterminate form \(\frac{\infty}{\infty}\).

### Step 2: Simplify the expression

To resolve the indeterminate form, we can divide the numerator and the denominator by \(\tan x\):

\[ = \lim_{x \to \frac{\pi}{2}} \sqrt{\frac{\frac{\tan x}{\tan x} - \frac{\sin(\tan^{-1}(\tan x))}{\tan x}}{\frac{\tan x}{\tan x} + \frac{\cos^2(\tan x)}{\tan x}}} \]

This simplifies to:

\[ = \lim_{x \to \frac{\pi}{2}} \sqrt{\frac{1 - \frac{\sin(\tan^{-1}(\tan x))}{\tan x}}{1 + \frac{\cos^2(\tan x)}{\tan x}}} \]

### Step 3: Evaluate the limit

As \( x \to \frac{\pi}{2} \):

- \(\frac{\sin(\tan^{-1}(\tan x))}{\tan x} \to \frac{1}{\infty} = 0\)
- \(\frac{\cos^2(\tan x)}{\tan x} \to \frac{\text{finite}}{\infty} = 0\)

Thus, we have:

\[ = \sqrt{\frac{1 - 0}{1 + 0}} = \sqrt{1} = 1 \]

### Final Answer

Therefore, the value of the limit is:

\[ \boxed{1} \]