Trigonometry & Inverse Trigonometry
tan⁻¹(4x) + tan⁻¹(6x) = π/6 — Number of Solutions
nta_pyq_2026_jan
Grade 12
Question:
The number of solutions of $\tan^{-1}4x+\tan^{-1}6x=\dfrac{\pi}{6}$, where $-\dfrac{1}{2\sqrt{6}}<x<\dfrac{1}{2\sqrt{6}}$, is equal to
Step-by-Step Solution
Key Concept: For $x$ in range, $24x^2<1$. Apply $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\tfrac{a+b}{1-ab}$: $\tan^{-1}\tfrac{10x}{1-24x^2}=\tfrac{\pi}{6}$. So $\tfrac{10x}{1-24x^2}=\tfrac{1}{\sqrt{3}}$.
1 solution.
Correct Answer: 4