Question

For what values of n, \( \tan^{-1}3 + \tan^{-1}n = \tan^{-1}\left(\frac{3+n}{1-3n}\right) \) is valid

• A. \( n \in (-\frac{1}{3}, \frac{1}{3}) \)
• B. \( n>\frac{1}{3} \)
• C. \( n<\frac{1}{3} \)
• D. For all real values of n

Hint:

Always remember to check the condition \(xy<1\) when applying the formula \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\). This is a common pitfall in trigonometry problems. The range of the principal value of \(\tan^{-1}\) is \( (-\pi/2, \pi/2) \), and this condition ensures the sum of the angles on the left also falls within a range that can be represented by a single \(\tan^{-1}\) function without adding or subtracting \(\pi\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The identity for the sum of two inverse tangent functions, \( \tan^{-1}x + \tan^{-1}y \), has different forms depending on the value of the product \(xy\). The principal value identity is given in the question.
Step 2: Key Formula or Approach:
The standard identity for the sum of inverse tangents is: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \] This formula is valid if and only if the product \(xy<1\).
If \(xy>1\) and \(x, y>0\), the formula becomes \( \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).
If \(xy>1\) and \(x, y<0\), the formula becomes \( -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).
Step 3: Detailed Explanation:
In the given equation, we have \(x = 3\) and \(y = n\). The equation is given in the standard form, which corresponds to the case where the principal value identity holds. The condition for this identity to be valid is: \[ xy<1 \] \[ 3n<1 \] Dividing by 3, we get: \[ n<\frac{1}{3} \] Step 4: Final Answer:
The given equation is valid for all values of \(n\) such that \(n<\frac{1}{3}\).