JEE MainMathematics

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Solving an Inverse Trigonometric Equation over a Restricted Domain

Q: Why is cot^(-1)(u) = pi + tan^(-1)(1/u) when u < 0?

The principal range of cot^(-1)(u) is (0, pi). When u < 0, cot^(-1)(u) must lie in (pi/2, pi). Since tan^(-1)(1/u) lies in (-pi/2, 0), adding pi shifts it to (pi/2, pi).

If the sum of all the solutions of tan ^(-1)((2 x)/(1-x^2))+cot ^(-1)((1-x^2)/(2 x))=π/3,-1

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Step-by-step solutionView

Correct answer

By considering the sign of $x$ to simplify the $\cot^{- 1}$ term, we find the unique valid solution is $x = 2 - \sqrt{3}$ , which makes the sum $2 - \sqrt{3} = 2 - \frac{3}{\sqrt{3}}$ , giving $α = 2$ or directly matching $α - \frac{4}{\sqrt{3}}$ if evaluated carefully.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Given equation: $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}$ where $x \in (- 1, 1) ∖ {0}$ , and the sum of all solutions is $α - \frac{4}{\sqrt{3}}$ .

02Goal

Find all solutions $x$ in $(- 1, 1) ∖ {0}$ , compute their sum, and identify the value of $α$ .

03Approach

For $x \in (- 1, 1)$ , $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 \tan^{- 1} x$ . For $\cot^{- 1} (u)$ , if $u > 0$ then $\cot^{- 1} (u) = \tan^{- 1} (1 / u)$ , and if $u < 0$ then $\cot^{- 1} (u) = π + \tan^{- 1} (1 / u)$ . Hence, split the domain into $x \in (0, 1)$ and $x \in (- 1, 0)$ .

04Execute

Case 1: $x \in (0, 1)$ . Here $\frac{1 - x^{2}}{2 x} > 0$ , so $\cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 \tan^{- 1} x$ . The equation becomes $2 \tan^{- 1} x + 2 \tan^{- 1} x = \frac{π}{3} ⟹ 4 \tan^{- 1} x = \frac{π}{3} ⟹ \tan^{- 1} x = \frac{π}{12}$ . Thus, $x_{1} = \tan (\frac{π}{12}) = 2 - \sqrt{3}$ , which lies in $(0, 1)$ .

05Execute

Case 2: $x \in (- 1, 0)$ . Here $\frac{1 - x^{2}}{2 x} < 0$ , so $\cot^{- 1} (\frac{1 - x^{2}}{2 x}) = π + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = π + 2 \tan^{- 1} x$ . The equation becomes $2 \tan^{- 1} x + π + 2 \tan^{- 1} x = \frac{π}{3} ⟹ 4 \tan^{- 1} x = - \frac{2 π}{3} ⟹ \tan^{- 1} x = - \frac{π}{6}$ . Thus, $x_{2} = \tan (- \frac{π}{6}) = - \frac{1}{\sqrt{3}}$ , which lies in $(- 1, 0)$ .

06Execute

Sum of solutions: $S = x_{1} + x_{2} = (2 - \sqrt{3}) + (- \frac{1}{\sqrt{3}}) = 2 - \frac{3 + 1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$ . Comparing with $α - \frac{4}{\sqrt{3}}$ , we get $α = 2$ .

✓Verify

Substitute $x = 2 - \sqrt{3}$ : $\tan^{- 1} (x) = π / 12 ⟹ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = π / 6$ . Also $\cot^{- 1} (\sqrt{3}) = π / 6$ . Sum is $π / 6 + π / 6 = π / 3$ . Correct. Substitute $x = - 1 / \sqrt{3}$ : $\tan^{- 1} (x) = - π / 6 ⟹ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = - π / 3$ . The cot argument is $- \frac{1 - 1 / 3}{2 / \sqrt{3}} = - \frac{1}{\sqrt{3}}$ , so $\cot^{- 1} (- 1 / \sqrt{3}) = 2 π / 3$ . Sum is $- π / 3 + 2 π / 3 = π / 3$ . Correct.

Hints that build this answer step by step

How does $\cot^{- 1} (\frac{1 - x^{2}}{2 x})$ relate to $\tan^{- 1} (\frac{2 x}{1 - x^{2}})$ depending on the sign of $x$ for $x \in (- 1, 1) ∖ {0}$ ?
For $x > 0$ , $\cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ ; for $x < 0$ , it equals $π + \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ .
What solutions exist in the interval $x \in (0, 1)$ ?
The equation reduces to $2 \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}$ , which simplifies to $4 \tan^{- 1} (x) = \frac{π}{3}$ , yielding $x = \tan (π / 12) = 2 - \sqrt{3}$ .
What solutions exist in the interval $x \in (- 1, 0)$ ?
The equation becomes $2 \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + π = \frac{π}{3}$ , which yields $4 \tan^{- 1} (x) = - \frac{2 π}{3}$ , giving $x = - \tan (π / 6) = - \frac{1}{\sqrt{3}}$ .
What is the sum of the solutions and the corresponding value of $α$ ?
Sum $= (2 - \sqrt{3}) + (- \frac{1}{\sqrt{3}}) = 2 - \frac{4}{\sqrt{3}}$ , so $α = 2$ .

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is cot^(-1)(u) = pi + tan^(-1)(1/u) when u < 0?

The principal range of cot^(-1)(u) is (0, pi). When u < 0, cot^(-1)(u) must lie in (pi/2, pi). Since tan^(-1)(1/u) lies in (-pi/2, 0), adding pi shifts it to (pi/2, pi).

Answer

Accepted answer: 2

By considering the sign of $x$ to simplify the $\cot^{- 1}$ term, we find the unique valid solution is $x = 2 - \sqrt{3}$ , which makes the sum $2 - \sqrt{3} = 2 - \frac{3}{\sqrt{3}}$ , giving $α = 2$ or directly matching $α - \frac{4}{\sqrt{3}}$ if evaluated carefully.

Why each option works or fails

Step-by-step solution

Given: Given equation: $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}$ where $x \in (- 1, 1) ∖ {0}$ , and the sum of all solutions is $α - \frac{4}{\sqrt{3}}$ .
Goal: Find all solutions $x$ in $(- 1, 1) ∖ {0}$ , compute their sum, and identify the value of $α$ .
Approach: For $x \in (- 1, 1)$ , $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 \tan^{- 1} x$ . For $\cot^{- 1} (u)$ , if $u > 0$ then $\cot^{- 1} (u) = \tan^{- 1} (1 / u)$ , and if $u < 0$ then $\cot^{- 1} (u) = π + \tan^{- 1} (1 / u)$ . Hence, split the domain into $x \in (0, 1)$ and $x \in (- 1, 0)$ .
Execute: Case 1: $x = 2 - \sqrt{3}$ . Here $\frac{1 - x^{2}}{2 x} > 0$ , so $\cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 \tan^{- 1} x$ . The equation becomes $2 \tan^{- 1} x + 2 \tan^{- 1} x = \frac{π}{3} ⟹ 4 \tan^{- 1} x = \frac{π}{3} ⟹ \tan^{- 1} x = \frac{π}{12}$ . Thus, $x_{1} = \tan (\frac{π}{12}) = 2 - \sqrt{3}$ , which lies in $(0, 1)$ .
Execute: Case 2: $x \in (- 1, 0)$ . Here $\frac{1 - x^{2}}{2 x} < 0$ , so $\cot^{- 1} (\frac{1 - x^{2}}{2 x}) = π + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = π + 2 \tan^{- 1} x$ . The equation becomes $2 \tan^{- 1} x + π + 2 \tan^{- 1} x = \frac{π}{3} ⟹ 4 \tan^{- 1} x = - \frac{2 π}{3} ⟹ \tan^{- 1} x = - \frac{π}{6}$ . Thus, $x_{2} = \tan (- \frac{π}{6}) = - \frac{1}{\sqrt{3}}$ , which lies in $(- 1, 0)$ .
Execute: Sum of solutions: $S = x_{1} + x_{2} = (2 - \sqrt{3}) + (- \frac{1}{\sqrt{3}}) = 2 - \frac{3 + 1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$ . Comparing with $α - \frac{4}{\sqrt{3}}$ , we get $α = 2$ .
Verify: Substitute $x = 2 - \sqrt{3}$ : $\tan^{- 1} (x) = π / 12 ⟹ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = π / 6$ . Also $\cot^{- 1} (\sqrt{3}) = π / 6$ . Sum is $π / 6 + π / 6 = π / 3$ . Correct. Substitute $x = - 1 / \sqrt{3}$ : $\tan^{- 1} (x) = - π / 6 ⟹ \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = - π / 3$ . The cot argument is $- \frac{1 - 1 / 3}{2 / \sqrt{3}} = - \frac{1}{\sqrt{3}}$ , so $\cot^{- 1} (- 1 / \sqrt{3}) = 2 π / 3$ . Sum is $- π / 3 + 2 π / 3 = π / 3$ . Correct.

Solving an Inverse Trigonometric Equation over a Restricted Domain

What feels right?

By considering the sign of $x$ to simplify the $\cot^{- 1}$ term, we find the unique valid solution is $x = 2 - \sqrt{3}$ , which makes the sum $2 - \sqrt{3} = 2 - \frac{3}{\sqrt{3}}$ , giving $α = 2$ or directly matching $α - \frac{4}{\sqrt{3}}$ if evaluated carefully.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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