Question

Evaluate: \(\tan \left( \tan^{-1} \frac{3}{1} + \tan^{-1} \frac{2}{1} \right)\)

• A. \(1\)
• B. \(0\)
• C. \(2\)
• D. \(3\)

Hint:

Use the tangent addition formula to simplify expressions involving the sum of inverse tangent functions: \[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} \]

Answer 1

The Correct Option is C

Step 1: Let \(\theta_1 = \tan^{-1} \frac{3}{1}\) and \(\theta_2 = \tan^{-1} \frac{2}{1}\). Thus, \[ \tan \theta_1 = 3 \quad \text{and} \quad \tan \theta_2 = 2 \] We need to evaluate: \[ \tan \left( \theta_1 + \theta_2 \right) \] Step 2: Use the tangent addition formula: \[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} \] Substitute the values: \[ \tan \left( \theta_1 + \theta_2 \right) = \frac{3 + 2}{1 - 3 \cdot 2} = \frac{5}{1 - 6} = \frac{5}{-5} = -1 \] Step 3: Hence, \[ \tan \left( \tan^{-1} \frac{3}{1} + \tan^{-1} \frac{2}{1} \right) = -1 \]