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Question #41171

Difficulty: 8/10 | 🟢 Free

The Question

Let $x = \tan \frac{\pi}{18}$ , $y = \tan \frac{2\pi}{9}$ , and $z = \tan \frac{7\pi}{18}$ . Define the matrix $M$ as: $M = \begin{bmatrix} \frac{2x}{1+x^2} & \frac{1-x^2}{1+x^2} & 1 \\\\ \frac{2y}{1+y^2} & \frac{1-y^2}{1+y^2} & 1 \\\\ \frac{2z}{1+z^2} & \frac{1-z^2}{1+z^2} & 1 \end{bmatrix}$ If $N = \text{adj}(M)$ , find the value of $\cos^{-1} \left( \frac{4}{3\sqrt{3}} \det(M N) \right)$ .

Options

A.
$\frac{\pi}{3}$
B.
$\frac{2\pi}{3}$
C.
$\frac{5\pi}{6}$
D.
$\frac{\pi}{6}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Variable Transformation Subsitution|Adjoint-Determinant Relationship|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate Inverse Trigonometric Function (ITF) substitutions with matrix determinant properties, specifically targeting the 'Row/Column Exchange Parity' trap.

Students commonly memorize the standard determinant form for $V = [\mathbf{1}, \mathbf{\cos}, \mathbf{\sin}]$ which yields a positive value for an increasing arithmetic progression of angles.

However, the given matrix $M = [\mathbf{\sin}, \mathbf{\cos}, \mathbf{1}]$ requires a single column swap ( $C_1 \leftrightarrow C_3$ ) to match this standard form, introducing a factor of $(-1)^1 = -1$ .

Failing to account for this parity inversion leads to $\det(M) = +\frac{\sqrt{3}}{2}$ , resulting in an incorrect final angle of $\frac{\pi}{3}$ instead of the true $\frac{2\pi}{3}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Variable Transformation Subsitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free