The value of $\lim_{x \to \frac{\pi}{2}} \sqrt{\frac{\tan x - \sin\left[\tan^{-1}(\tan x)\right]}{\tan x + \cos^2(\tan x)}}$ is______.
Step-by-Step Solution
Key Concept: Recognizing that $\tan^{-1}(\tan x) = x - \pi$ near $x = \pi/2$ transforms the limit into an evaluable form.
We evaluate $\lim_{x \to \pi/2} \frac{\tan x - \sin[\tan^{-1}(\tan x)]}{\tan x + \cos^2(\tan x)}$. Since $\tan^{-1}(\tan x) = x - \pi$ when $x \to \pi/2^-$, the numerator becomes $\tan x - \sin(x - \pi) = \tan x + \sin x$. Computing the limit using algebraic manipulation and simplification, we find $\lim_{x \to \pi/2^-} \frac{\tan x + \sin x}{\tan x + \cos^2(\tan x)} = 1$.
Correct Answer: 1200