Trigonometry & Inverse Trigonometry
Solving equations involving inverse trig functions
nta_pyq_2023_jan
Grade 12
Question:
If the sum of all the solutions of \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}, \quad -1 < x < 1, x \neq 0, \text{ is } \alpha - \frac{4}{\sqrt{3}}, \text{ then } \alpha \text{ is equal to}
Step-by-Step Solution
Key Concept: Use the identity \tan^{-1}(t) + \cot^{-1}(t) = \pi/2 and split into cases x > 0 and x < 0.
Case I (x > 0): equation reduces to 2\tan^{-1}x = \pi/3, giving x = 2-\sqrt{3}. Case II (x < 0): 2\tan^{-1}x + \pi = \pi/3, giving x = -1/\sqrt{3}. Sum = (2-\sqrt{3}) + (-1/\sqrt{3}) = 2 - \sqrt{3} - 1/\sqrt{3} = 2 - 4/\sqrt{3}. So \alpha = 2.
Correct Answer: 2