Trigonometry & Inverse Trigonometry
Equation involving inverse tan and inverse cos
nta_pyq_2023_jan
Grade None
Question:
Let S = \left\{x \in \mathbb{R} : 0 < x < 1 \text{ and } 2\tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}. If n(S) denotes the number of elements in S then:
n(S) = 2 and only one element in S is less than \frac{1}{2}
n(S) = 1 and the element in S is more than \frac{1}{2}
n(S) = 1 and the element in S is less than \frac{1}{2}
n(S) = 0
Step-by-Step Solution
Key Concept: Let \tan^{-1}x = \theta, then simplify using double-angle formulas. Solve 4\theta = \pi/2.
For 0 < x < 1, let \tan^{-1}x = \theta \in (0, \pi/4). LHS = 2\tan^{-1}(\tan(\pi/4 - \theta)) = 2(\pi/4 - \theta). RHS = \cos^{-1}(\cos 2\theta) = 2\theta. So \pi/2 - 2\theta = 2\theta, \theta = \pi/8. x = \tan(\pi/8) = \sqrt{2}-1 \approx 0.414 < 1/2. n(S) = 1.
Correct Answer: 3