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 $\frac{17}{6}$.