JEE Main 2023 (Online) 24th January Evening ShiftQuestion 1
Let $S=\{\theta\in[0,2\pi): \tan(\pi\cos\theta)+\tan(\pi\sin\theta)=0\}$, then $\sum_{\theta\in S}\sin^{2}\left(\theta+\frac\pi4\right)$ is equal to
Trigonometric Ratio & Identities
Answer$2$
Solution
Concept. $\tan(\pi\cos\theta)=\tan(-\pi\sin\theta)$ gives $\sin\theta+\cos\theta$ equal to an integer. Formulas used. (1) $\tan x=\tan y\iff x=y+k\pi$(2) $-\sqrt2\le\sin\theta+\cos\theta\le\sqrt2$Step 1 - Equation. By (1), $\pi\cos\theta=-\pi\sin\theta+k\pi$, so $\sin\theta+\cos\theta=k$. By (2), $k\in\{-1,0,1\}$. Step 2 - Solutions in $[0,2\pi)$. $k=1$: $\theta=0,\frac\pi2$. $k=-1$: $\theta=\pi,\frac{3\pi}2$. $k=0$: $\theta=\frac{3\pi}4,\frac{7\pi}4$. At all of these both tangents are defined (for $k=0$ the angles are $\pm\frac{\pi}{\sqrt2}$). Step 3 - Sum. $\sin^2\left(\theta+\frac\pi4\right)$ is $\frac12$ at each of $0,\frac\pi2,\pi,\frac{3\pi}2$ and $0$ at $\frac{3\pi}4,\frac{7\pi}4$. Total $4\cdot\frac12=2$. Answer: 2