sin 2 theta+tan 2 theta0 થાય તેવી છે thet | vedclass

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Question

$\sin 2 	heta+	an 2 	heta>0$

$\Rightarrow \sin 2 	heta+\frac{\sin 2 	heta}{\cos 2 	heta}>0$

$\Rightarrow \sin 2 	heta \frac{(\cos 2

Vedclass · Accepted Answer

$\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right) \cup\left(\pi, \frac{5 \pi}{4}\right) \cup\left(\frac{3 \pi}{2}, \frac{7 \pi}{4}\right)$

Vedclass · Answer

$\sin 2 	heta+	an 2 	heta>0$

$\Rightarrow \sin 2 	heta+\frac{\sin 2 	heta}{\cos 2 	heta}>0$

$\Rightarrow \sin 2 	heta \frac{(\cos 2 	heta+1)}{\cos 2 	heta}>0 \Rightarrow 	an 2 	heta\left(2 \cos ^{2} 	hetaight)>0$

Note : $\cos 2 	heta 
eq 0$

$\Rightarrow 1-2 \sin ^{2} 	heta 
eq 0 \Rightarrow \sin 	heta 
eq \pm \frac{1}{\sqrt{2}}$

Now, $	an 2 	heta(1+\cos 2 	heta)>0$

$\Rightarrow 	an 2 	heta>0 \quad($ as $\cos 2 	heta+1>0)$

$\Rightarrow 2 	heta \in\left(0, \frac{\pi}{2}ight) \cup\left(\pi, \frac{3 \pi}{2}ight) \cup\left(2 \pi, \frac{5 \pi}{2}ight) \cup\left(3 \pi, \frac{7 \pi}{2}ight)$

$\Rightarrow 	heta \in\left(0, \frac{\pi}{4}ight) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{4}ight) \cup\left(\pi, \frac{5 \pi}{4}ight) \cup\left(\frac{3 \pi}{2}, \frac{7 \pi}{4}ight)$

As $\sin 	heta 
eq \pm \frac{1}{\sqrt{2}} ;$ which has been already considered

sin $2 \theta+\tan 2 \theta>0$ થાય તેવી છે $\theta \in[0,2 \pi]$ ની શક્ય તમામ કિંમતો ........... માં આપેલ છે.

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