Question

The value of \( \tan \left[ \tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \left( \frac{2}{3} \right) \right] \) is:

• A. \( \frac{17}{6} \)
• B. \( \frac{6}{17} \)
• C. \( -\frac{17}{6} \)
• D. \( -\frac{6}{11} \)
• E. 1

Hint:

Use the sum of inverse tangent formula to simplify expressions with multiple inverse tangents.

Answer 1

The Correct Option is A

Using the identity for the sum of inverse tangents, we have: \[ \tan \left( \tan^{-1} a + \tan^{-1} b \right) = \frac{a + b}{1 - ab} \] Substituting \( a = \frac{3}{4} \) and \( b = \frac{2}{3} \): \[ \tan \left( \tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \left( \frac{2}{3} \right) \right) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left( \frac{3}{4} \times \frac{2}{3} \right)} \] Simplifying the numerator and denominator: \[ = \frac{\frac{9}{12} + \frac{8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{6}{12}} = \frac{17}{6} \] Thus, the value is \( \frac{17}{6} \).