Question:
If
\[
\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4},
\]
then find the value of \(x\):
If
\[
\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4},
\]
then find the value of \(x\):
Show Hint
Whenever an equation contains a sum of inverse tangent functions,
apply the tangent addition formula and reduce the equation to an
algebraic form.
Updated On: Jun 11, 2026
- \(2\)
- \(1\)
- \(\boxed{x=\frac16}\)
- \(0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For inverse tangent functions,
\[
\tan(A+B)
=
\frac{\tan A+\tan B}
{1-\tan A\tan B}.
\]
This identity is used to convert the equation into a quadratic equation.
Step 1: Apply tangent on both sides.
Given, \[ \tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}. \] Taking tangent, \[ \tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right) = \tan\frac{\pi}{4}. \] Since \[ \tan\frac{\pi}{4}=1, \] we get \[ \frac{3x+2x}{1-6x^2}=1. \]
Step 2: Form the quadratic equation.
\[ 5x=1-6x^2. \] Therefore, \[ 6x^2+5x-1=0. \] Factorizing, \[ (6x-1)(x+1)=0. \] Hence, \[ x=\frac16 \quad \text{or} \quad x=-1. \]
Step 3: Check the valid solution.
For \(x=-1\), \[ \tan^{-1}(-3)+\tan^{-1}(-2) \] is negative and cannot equal \[ \frac{\pi}{4}. \] Therefore, \[ \boxed{x=\frac16}. \]
Step 1: Apply tangent on both sides.
Given, \[ \tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}. \] Taking tangent, \[ \tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right) = \tan\frac{\pi}{4}. \] Since \[ \tan\frac{\pi}{4}=1, \] we get \[ \frac{3x+2x}{1-6x^2}=1. \]
Step 2: Form the quadratic equation.
\[ 5x=1-6x^2. \] Therefore, \[ 6x^2+5x-1=0. \] Factorizing, \[ (6x-1)(x+1)=0. \] Hence, \[ x=\frac16 \quad \text{or} \quad x=-1. \]
Step 3: Check the valid solution.
For \(x=-1\), \[ \tan^{-1}(-3)+\tan^{-1}(-2) \] is negative and cannot equal \[ \frac{\pi}{4}. \] Therefore, \[ \boxed{x=\frac16}. \]
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