JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
All possible values of $\theta\in[0,2\pi]$ for which $\sin2\theta+\tan2\theta>0$ lie in :
Trigonometric Ratio & Identities
Solution
Concept. $\sin2\theta+\tan2\theta=\tan2\theta\,(1+\cos2\theta)=2\cos^2\theta\tan2\theta$; the sign is the sign of $\tan2\theta$. Formulas used. (1) $\sin2\theta=\tan2\theta\cos2\theta$ and $1+\cos2\theta=2\cos^2\theta\ge0$Step 1 - Factor. By (1), we need $\tan2\theta>0$ and $\cos\theta\ne0$. Step 2 - Where $\tan2\theta>0$. For $2\theta\in[0,4\pi]$: $$2\theta\in\left(0,\tfrac\pi2\right)\cup\left(\pi,\tfrac{3\pi}2\right)\cup\left(2\pi,\tfrac{5\pi}2\right)\cup\left(3\pi,\tfrac{7\pi}2\right).$$ Step 3 - Divide by $2$. $$\theta\in\left(0,\tfrac\pi4\right)\cup\left(\tfrac\pi2,\tfrac{3\pi}4\right)\cup\left(\pi,\tfrac{5\pi}4\right)\cup\left(\tfrac{3\pi}2,\tfrac{7\pi}4\right).$$ Answer: (D)