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$\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}=$
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Verified Answer
The correct answer is:
$\tan 60$
$\begin{aligned}
& \frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta} \\
& =\frac{(\sin 3 \theta+\sin 9 \theta)+(\sin 5 \theta+\sin 7 \theta)}{(\cos 3 \theta+\cos 9 \theta)+(\cos 5 \theta+\cos 7 \theta} \\
& =\frac{2 \sin 6 \theta \cos 3 \theta+2 \sin 6 \theta \cos \theta}{2 \cos 6 \theta \cos 3 \theta+2 \cos 6 \theta \cos \theta} \\
& =\frac{2 \sin 6 \theta(\cos 3 \theta+\cos \theta)}{2 \cos 6 \theta(\cos 3 \theta+\cos \theta)}=\tan 6 \theta \\
&
\end{aligned}$
& \frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta} \\
& =\frac{(\sin 3 \theta+\sin 9 \theta)+(\sin 5 \theta+\sin 7 \theta)}{(\cos 3 \theta+\cos 9 \theta)+(\cos 5 \theta+\cos 7 \theta} \\
& =\frac{2 \sin 6 \theta \cos 3 \theta+2 \sin 6 \theta \cos \theta}{2 \cos 6 \theta \cos 3 \theta+2 \cos 6 \theta \cos \theta} \\
& =\frac{2 \sin 6 \theta(\cos 3 \theta+\cos \theta)}{2 \cos 6 \theta(\cos 3 \theta+\cos \theta)}=\tan 6 \theta \\
&
\end{aligned}$
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