Question
Find the number of solutions of:
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \]
Solution
Use identity:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
So,
\[ \tan^{-1}\left(\frac{2x + 3x}{1 – 6x^2}\right) = \frac{\pi}{4} \]
\[ \tan^{-1}\left(\frac{5x}{1 – 6x^2}\right) = \frac{\pi}{4} \]
Thus,
\[ \frac{5x}{1 – 6x^2} = 1 \]
\[ 5x = 1 – 6x^2 \]
\[ 6x^2 + 5x – 1 = 0 \]
Solve quadratic:
\[ x = \frac{-5 \pm \sqrt{25 + 24}}{12} = \frac{-5 \pm 7}{12} \]
\[ x = \frac{1}{6}, \quad x = -1 \]
Check validity:
- For \( x = \frac{1}{6} \): valid
- For \( x = -1 \): LHS becomes negative, cannot equal \( \frac{\pi}{4} \)
So only one valid solution.
Final Answer:
\[ \boxed{1} \]
Key Concept
Always verify solutions due to principal value restrictions of inverse tangent.