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Period of \(\frac{\sin \theta+\sin 2 \theta}{\cos \theta+\cos 2 \theta}\) is
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Verified Answer
The correct answer is:
\(\frac{2 \pi}{3}\)
\(\begin{aligned}
& \frac{\sin \theta+\sin 2 \theta}{\cos \theta+\cos 2 \theta}=\frac{2 \sin \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}=\tan \left(\frac{3 \theta}{2}\right) \\
& \text {Hence period }=\frac{2 \pi}{3}
\end{aligned}\)
& \frac{\sin \theta+\sin 2 \theta}{\cos \theta+\cos 2 \theta}=\frac{2 \sin \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}=\tan \left(\frac{3 \theta}{2}\right) \\
& \text {Hence period }=\frac{2 \pi}{3}
\end{aligned}\)
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