Question

The value of the following expression: $ \cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} + 1}{\tan 2} \right) - \cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} - 1}{\tan 2} \right) $

• A. \( \frac{\pi}{2} + \frac{5}{2} \)
• B. \( \frac{\pi}{2} - \frac{3}{2} \)
• C. \( 2 - \frac{\pi}{2} \)
• D. \( 3 + \frac{\pi}{2} \)

Hint:

For problems involving inverse cotangents, use the identity for the difference of cotangents to simplify the expressions.

Answer 1

The Correct Option is C

The provided expression is:\[\cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} + 1}{\tan 2} \right) - \cot^{-1} \left( \frac{\sqrt{1 + \tan^2 2} - 1}{\tan 2} \right)\]Using the identity \( \sqrt{1 + \tan^2 \theta} = \sec \theta \), the expression simplifies to:\[\cot^{-1} \left( \frac{\sec 2 + 1}{\tan 2} \right) - \cot^{-1} \left( \frac{\sec 2 - 1}{\tan 2} \right)\]Applying the cotangent difference identity \( \cot^{-1} A - \cot^{-1} B = \cot^{-1} \left( \frac{B - A}{1 + AB} \right) \), we get:\[\cot^{-1} \left( \frac{\frac{\sec 2 - 1}{\tan 2} - \frac{\sec 2 + 1}{\tan 2}}{1 + \frac{\sec 2 + 1}{\tan 2} \cdot \frac{\sec 2 - 1}{\tan 2}} \right)\]The simplified result is:\[2 - \frac{\pi}{2}\]Therefore, the correct answer is \( 2 - \frac{\pi}{2} \).