Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 12
Question:
Consider the equation $\tan^{-1}x + \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$. Let $\alpha =$ sum of positive integral solutions of $x$ and $\beta =$ sum of positive integral solutions of $y$. Then $\beta - \alpha$ = _______.
Step-by-Step Solution
Key Concept: Using inverse trigonometric identities to convert the equation into a linear Diophantine form allows systematic enumeration of integer solutions.
Starting from $\tan^{-1}x + \cos^{-1}\frac{2}{\sqrt{1+x^2}} = \sin^{-1}\frac{3}{\sqrt{10}}$, we use the identity $\tan^{-1}x + \tan^{-1}\frac{1}{x} = \tan^{-1}(3)$ to find $\frac{xy+1}{y-x} = 3$. This yields $xy + 1 = 3y - 3x$ or $y = \frac{3x+1}{3-x}$. The positive integral solutions are $(1,2)$ and $(2,7)$, giving $\alpha = 1 + 2 = 3$ and $\beta = 2 + 7 = 9$, so $\beta - \alpha = 6$.
Correct Answer: 6