Question

If \(tan^{-1}ax+tan^{-1}3x = \frac{π}{4}\), where \(3ax^2 < 1\), then value of \(a\) for \(x = \frac{1}{6}\) is \(\ldots\)

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(9\)

Hint:

Use tan^-1 A + tan^-1 B = tan^-1((A+B)/(1-AB)) and put x = 1/6.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The sum of two inverse tangents can be combined into one when the product of the arguments is less than 1, which the condition \(3ax^2 < 1\) guarantees.

Step 2: Key Formula or Approach:
\[ \tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A + B}{1 - AB}, \quad AB < 1 \]

Step 3: Detailed Explanation:
Here \(A = ax\) and \(B = 3x\):
\[ \tan^{-1}\frac{ax + 3x}{1 - 3ax^2} = \frac{\pi}{4} \]
Take tangent on both sides, using \(\tan\frac{\pi}{4} = 1\):
\[ \frac{x(a+3)}{1 - 3ax^2} = 1 \]
Substitute \(x = \tfrac16\), so \(x^2 = \tfrac{1}{36}\):
\[ \frac{(a+3)/6}{1 - a/12} = 1 \Rightarrow \frac{a+3}{6} = 1 - \frac{a}{12} \]
Multiply by 12:
\[ 2(a + 3) = 12 - a \Rightarrow 3a = 6 \Rightarrow a = 2 \]
Check: \(\tan^{-1}\tfrac13 + \tan^{-1}\tfrac12 = \tan^{-1}\dfrac{5/6}{5/6} = \tan^{-1}1 = \tfrac{\pi}{4}\). It holds.

Final Answer:
\(a = 2\), option (A). \[ \boxed{2 \text{ (A)}} \]