Question

The value of \[ \tan\!\left(\tan^{-1}(3)+\tan^{-1}(7)\right) \] is equal to:

• A. \(-\frac{1}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{5}\)
• D. \(-\frac{1}{5}\)
• E. 0

Hint:

Since \(3 \times 7 > 1\), the angle \(\tan^{-1}(3) + \tan^{-1}(7)\) will be greater than \(\frac{\pi}{2}\). This explains why the tangent value is negative (it falls in the second quadrant).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the addition formula for the tangent function applied to inverse tangent arguments.

Step 2: Key Formula or Approach:
Formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).
Here \(A = \tan^{-1}(3)\) and \(B = \tan^{-1}(7)\), so \(\tan A = 3\) and \(\tan B = 7\).

Step 3: Detailed Explanation:
Substitute the values into the formula:
\[ \text{Value} = \frac{3 + 7}{1 - (3 \times 7)} \]
\[ \text{Value} = \frac{10}{1 - 21} \]
\[ \text{Value} = \frac{10}{-20} \]
\[ \text{Value} = -\frac{1}{2} \]

Step 4: Final Answer:
The value is \(-\frac{1}{2}\).