JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
The maximum value of $(\cos\theta_1)\cdot(\cos\theta_2)\cdot(\cos\theta_3)\cdot\ldots\cdot(\cos\theta_{10})$ under the restrictions $0<\theta_1,\theta_2,\ldots,\theta_{10}<\frac{\pi}{2}$ and $(\cot\theta_1)\cdot(\cot\theta_2)\cdot\ldots\cdot(\cot\theta_{10})=1$ is
Trigonometric Ratio & Identities
Solution
Concept. $\prod\cot\theta_i=1$ means $\prod\cos\theta_i=\prod\sin\theta_i$; multiply them. Formulas used. (1) $\sin\theta\cos\theta=\frac12\sin2\theta\le\frac12$Step 1 - Square. Let $P=\prod\cos\theta_i=\prod\sin\theta_i$. By (1), $$P^2=\prod\sin\theta_i\cos\theta_i=\frac{\prod\sin2\theta_i}{2^{10}}\le\frac{1}{2^{10}}.$$ Step 2 - Maximum. $P\le\frac{1}{32}$, with equality at $\theta_i=\frac\pi4$ (which satisfies the condition). Answer: (D)