JEE PYQ: Inverse Trigonometric Functions Question 4
Question 4 - 2021 (17 Mar Shift 1)
The sum of possible values of $x$ for $\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\left(\frac{8}{31}\right)$ is:
(1) $-\frac{32}{4}$
(2) $-\frac{31}{4}$
(3) $-\frac{30}{4}$
(4) $-\frac{33}{4}$
Show Answer
Answer: (1)
Solution
Taking tangent both sides: $\frac{(x+1)+(x-1)}{1-(x^2-1)} = \frac{8}{31}$. So $\frac{2x}{2-x^2} = \frac{8}{31}$. This gives $4x^2 + 31x - 8 = 0$, so $x = -8$ or $x = \frac{1}{4}$. But if $x = \frac{1}{4}$, then $\tan^{-1}(x+1) \in \left(0, \frac{\pi}{2}\right)$ and $\cot^{-1}\left(\frac{1}{x-1}\right) \in \left(\frac{\pi}{2}, \pi\right)$, so LHS $> \frac{\pi}{2}$ while RHS $< \frac{\pi}{2}$. Not possible. Hence $x = -8$, and sum = $-\frac{32}{4}$.
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