Question

\(\tan^{-1}[\frac{1}{\sqrt3}\sin\frac{5\pi}{2}]\sin^{-1}[\cos(\sin^{-1}\frac{\sqrt3}{2})]=\)

• A. 0
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{6}\)
• D. π

Answer 1

The Correct Option is A, B, C, D

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 \(\frac{\pi}{3}\)