Question

Evaluate \( \sin \left( \tan^{-1}\frac{4}{5} + \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{9} - \tan^{-1}\frac{1}{7} \right) \):

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( 1 \)

Hint:

For trigonometric sums involving inverse functions, use addition and subtraction formulas to simplify the expressions before evaluating.

Answer 1

The Correct Option is D

The task is to evaluate: \[ \sin\left(\tan^{-1}\frac{4}{5} + \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{9} - \tan^{-1}\frac{1}{7}\right). \] Step 1: Use the arctangent addition formula to simplify. The formula for the sum of two arctangents is: \[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right), \quad \text{provided } xy<1. \] Apply this formula to pairs of terms: \[ \tan^{-1}\frac{4}{5} + \tan^{-1}\frac{1}{9} = \tan^{-1}\left(\frac{\frac{4}{5} + \frac{1}{9}}{1 - \frac{4}{5} \cdot \frac{1}{9}}\right). \] Simplify the expression: \[ \tan^{-1}\left(\frac{\frac{36}{45} + \frac{5}{45}}{1 - \frac{4}{45}}\right) = \tan^{-1}\left(\frac{\frac{41}{45}}{\frac{41}{45}}\right) = \tan^{-1}(1). \] Similarly, for the other pair: \[ \tan^{-1}\frac{4}{3} - \tan^{-1}\frac{1}{7} = \tan^{-1}\left(\frac{\frac{4}{3} - \frac{1}{7}}{1 + \frac{4}{3} \cdot \frac{1}{7}}\right). \] Simplify: \[ \tan^{-1}\left(\frac{\frac{28}{21} - \frac{3}{21}}{1 + \frac{4}{21}}\right) = \tan^{-1}\left(\frac{\frac{25}{21}}{\frac{25}{21}}\right) = \tan^{-1}(1). \] Step 2: Combine the simplified results. The expression now becomes: \[ \sin\left(\tan^{-1}(1) + \tan^{-1}(1)\right). \] Step 3: Perform further simplification. Since \(\tan^{-1}(1) = \frac{\pi}{4}\), substitute this value: \[ \sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right). \] As \(\sin\left(\frac{\pi}{2}\right) = 1\), the final result is: \[ \boxed{1}. \]