We need to evaluate the expression:
\[ \tan^{-1}\left[\frac{1}{\sqrt{3}} \sin\left(\frac{5\pi}{2}\right)\right] \sin^{-1}\left[\cos\left(\sin^{-1}\frac{\sqrt{3}}{2}\right)\right] \]
Let's evaluate the two main parts of the expression separately.
Part 1: \( \tan^{-1}\left[\frac{1}{\sqrt{3}} \sin\left(\frac{5\pi}{2}\right)\right] \)
First, evaluate \( \sin\left(\frac{5\pi}{2}\right) \):
\[ \frac{5\pi}{2} = \frac{4\pi + \pi}{2} = 2\pi + \frac{\pi}{2} \]
Since \( \sin(x) \) has a period of \( 2\pi \):
\[ \sin\left(\frac{5\pi}{2}\right) = \sin\left(2\pi + \frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
Now substitute this back into the first part:
\[ \tan^{-1}\left[\frac{1}{\sqrt{3}} \times 1\right] = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
The principal value range for \( \tan^{-1}(x) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). The angle \( \theta \) in this range such that \( \tan(\theta) = \frac{1}{\sqrt{3}} \) is \( \frac{\pi}{6} \).
So, Part 1 evaluates to \( \frac{\pi}{6} \).
Part 2: \( \sin^{-1}\left[\cos\left(\sin^{-1}\frac{\sqrt{3}}{2}\right)\right] \)
First, evaluate the innermost term \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \):
The principal value range for \( \sin^{-1}(x) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). The angle \( \phi \) in this range such that \( \sin(\phi) = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \).
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]
Now substitute this back into the second part:
\[ \sin^{-1}\left[\cos\left(\frac{\pi}{3}\right)\right] \]
Evaluate \( \cos\left(\frac{\pi}{3}\right) \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Substitute this back:
\[ \sin^{-1}\left(\frac{1}{2}\right) \]
The angle \( \psi \) in the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) such that \( \sin(\psi) = \frac{1}{2} \) is \( \frac{\pi}{6} \).
So, Part 2 evaluates to \( \frac{\pi}{6} \).
Combining the Parts:
The original expression shows the two parts written next to each other: \( \left(\frac{\pi}{6}\right) \left(\frac{\pi}{6}\right) \). In standard mathematical notation, juxtaposition implies multiplication.
\[ \left(\frac{\pi}{6}\right) \times \left(\frac{\pi}{6}\right) = \frac{\pi^2}{36} \]
However, \( \frac{\pi^2}{36} \) is not among the options (A) 0, (B) \( \frac{\pi}{6} \), (C) \( \frac{\pi}{3} \), (D) \( \pi \).
This suggests there might be a typo in the question, possibly an omitted operator. If we assume the intended operation between the two parts was addition instead of multiplication:
\[ \text{Part 1} + \text{Part 2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]
This result, \( \frac{\pi}{3} \), matches option (C).
Assuming addition was intended due to the options provided:
The final answer is \(\frac{\pi}{3}\)
The range of the real valued function \( f(x) =\) \(\sin^{-1} \left( \frac{1 + x^2}{2x} \right)\) \(+ \cos^{-1} \left( \frac{2x}{1 + x^2} \right)\) is: