JEE Main 2021Trigonometric EquationsTrigonometric EquationsMediumQuestionAll possible values of θ\thetaθ ∈\in∈ [0, 2π\piπ] for which sin 2θ\thetaθ + tan 2θ\thetaθ > 0 lie in :OptionsA(0,π4)∪(π2,3π4)∪(3π2,11π6)\left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{11\pi } \over 6}} \right)(0,4π)∪(2π,43π)∪(23π,611π)B(0,π2)∪(π,3π2)\left( {0,{\pi \over 2}} \right) \cup \left( {\pi ,{{3\pi } \over 2}} \right)(0,2π)∪(π,23π)C(0,π2)∪(π2,3π4)∪(π,7π6)\left( {0,{\pi \over 2}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{7\pi } \over 6}} \right)(0,2π)∪(2π,43π)∪(π,67π)D(0,π4)∪(π2,3π4)∪(π,5π4)∪(3π2,7π4)\left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{5\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{7\pi } \over 4}} \right)(0,4π)∪(2π,43π)∪(π,45π)∪(23π,47π)Check AnswerHide SolutionSolutionsin2θ+tan2θ>0\sin 2\theta + \tan 2\theta > 0sin2θ+tan2θ>0 ⇒sin2θ+sin2θcos2θ>0 \Rightarrow \sin 2\theta + {{\sin 2\theta } \over {\cos 2\theta }} > 0⇒sin2θ+cos2θsin2θ>0 ⇒sin2θ(cos2θ+1)cos2θ>0⇒tan2θ(2cos2θ)>0 \Rightarrow \sin 2\theta {{(\cos 2\theta + 1)} \over {\cos 2\theta }} > 0 \Rightarrow \tan 2\theta (2{\cos ^2}\theta ) > 0⇒sin2θcos2θ(cos2θ+1)>0⇒tan2θ(2cos2θ)>0 Note : cos2θ≠0\cos 2\theta \ne 0cos2θ=0 ⇒1−2sin2θ≠θ⇒sinθ≠±12 \Rightarrow 1 - 2{\sin ^2}\theta \ne \theta \Rightarrow \sin \theta \ne \pm {1 \over {\sqrt 2 }}⇒1−2sin2θ=θ⇒sinθ=±21 Now, tan2θ(1+cos2θ)>0\tan 2\theta (1 + \cos 2\theta ) > 0tan2θ(1+cos2θ)>0 ⇒tan2θ>0 \Rightarrow \tan 2\theta > 0⇒tan2θ>0 (as cos2θ+1>0\cos 2\theta + 1 > 0cos2θ+1>0) ⇒2θ∈(0,π2)∪(π,3π2)∪(2π,5π2)∪(3π,7π2) \Rightarrow 2\theta \in \left( {0,{\pi \over 2}} \right) \cup \left( {\pi ,{{3\pi } \over 2}} \right) \cup \left( {2\pi ,{{5\pi } \over 2}} \right) \cup \left( {3\pi ,{{7\pi } \over 2}} \right)⇒2θ∈(0,2π)∪(π,23π)∪(2π,25π)∪(3π,27π) ⇒θ∈(0,π4)∪(π2,3π4)∪(π,5π4)∪(3π2,7π4) \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{5\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{7\pi } \over 4}} \right)⇒θ∈(0,4π)∪(2π,43π)∪(π,45π)∪(23π,47π)