If 0, then the expression ^-1 + ( 1 + ^2 -) + ^-1 + ( 1 + ^2 -) + ^-1 + ( 1 + ^2 -) is equal to:

Given: 0 and the expression ^-1 + 1 + ^2 - + ^-1 + 1 + ^2 - + ^-1 + 1 + ^2 - Find: Its value. The solution states that the given expression involves inverse cotangents and algebraic manipulation. By applying trigonometric identities and simplifying, the expression reduces to. Therefore, the correct option is D, and the value of the expression is. Given: 0. Find: The value of ^-1 + 1 + ^2 - + ^-1 + 1 + ^2 - + ^-1 + 1 + ^2 - From the extracted solution: 1. Understand the given expression. 2. Identify patterns involving inverse cotangent sums. 3. Apply the identity ^-1 x + ^-1 y = ^-1 ( xy - 1x + y) 4. Use algebraic manipulation together with the order 0. The solution concludes that after simplification, the expression evaluates to Thus, the correct option is D.

If 0, then the expression ^-1 + ( 1 + ^2 -) + ^-1 + ( 1 + ^2 -) +… — Solution | JEE Mathematics

MCQMediumJEE 2025Inverse Trigonometric Functions

If $\alpha > \beta > \gamma > 0$ , then the expression $\cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right)$ is equal to:

A
$3\pi$
B
$\frac{\pi}{2} - (\alpha + \beta + \gamma)$
C
$0$
D
$\pi$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $\alpha > \beta > \gamma > 0$ and the expression

\cot^{-1} \beta + \frac{1 + \beta^2}{\alpha - \beta} + \cot^{-1} \gamma + \frac{1 + \gamma^2}{\beta - \gamma} + \cot^{-1} \alpha + \frac{1 + \alpha^2}{\gamma - \alpha}

Find: Its value.

The solution states that the given expression involves inverse cotangents and algebraic manipulation. By applying trigonometric identities and simplifying, the expression reduces to $\pi$ .

Therefore, the correct option is D, and the value of the expression is $\pi$ .

Identity-Based Approach

Given: $\alpha > \beta > \gamma > 0$ .

Find: The value of

\cot^{-1} \beta + \frac{1 + \beta^2}{\alpha - \beta} + \cot^{-1} \gamma + \frac{1 + \gamma^2}{\beta - \gamma} + \cot^{-1} \alpha + \frac{1 + \alpha^2}{\gamma - \alpha}

From the extracted solution:

Understand the given expression.
Identify patterns involving inverse cotangent sums.
Apply the identity

\cot^{-1} x + \cot^{-1} y = \cot^{-1} \left( \frac{xy - 1}{x + y} \right)

Use algebraic manipulation together with the order $\alpha > \beta > \gamma > 0$ .

The solution concludes that after simplification, the expression evaluates to

\pi

Thus, the correct option is D.

Common mistakes

Treating the fractional terms as independent algebraic add-ons without looking for an inverse trigonometric identity is incorrect. The expression is designed to be simplified through a pattern involving $\cot^{-1}$ terms. First search for a standard identity before expanding anything.
Using the inverse cotangent sum identity without checking the domain conditions can give a wrong branch of the angle. Since $\alpha > \beta > \gamma > 0$ , the ordering helps determine the correct principal value. Always use the given inequalities while fixing the final angle.
Assuming the answer must be $0$ because of apparent cyclic symmetry is misleading. Cyclic expressions in inverse trigonometric functions often differ by a constant such as $\pi$ because of principal value conventions. Verify the final value after simplification instead of relying only on symmetry.

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Answer

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