Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
If the sum of all values of $\theta$, $0 \leq \theta \leq 2\pi$ satisfying the equation $(8\cos 40 - 3)(\cot \theta + \tan \theta - 2)(\cot \theta + \tan \theta + 2) = 12$ is $k\pi$, then $k$ is equal to:
Step-by-Step Solution
Key Concept: Expressing cotangent-tangent differences using double-angle formulas reveals hidden polynomial structure.
The equation $(8\cos 40 - 3)(\cot \theta - \tan \theta)^2 = 12$ simplifies using $\cot \theta - \tan \theta = \frac{4\cos^2 2\theta}{\sin^2 2\theta}$ to $16\cos^2 2\theta - 8\cos^2 2\theta - 3 = 0$. Factoring gives $4\cos^2 2\theta - 3)(4\cos^2 2\theta + 1) = 0$, yielding $\cos 2\theta = \pm\frac{\sqrt{3}}{2}$.
Correct Answer: 8