$\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}=$

$\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}=$
  1. $\cot 7 A$
  2. $\tan 6 A$
  3. $\tan 7 A$
  4. $\cot 6 A$

Solution

$\begin{aligned} \frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A} &=\frac{(\sin A+\sin 13 A)+\sin 7 A}{(\cos A+\cos 13 A)+\cos 7 A} \\ &=\frac{2 \sin 7 A \cos 6 A+\sin 7 A}{2 \cos 7 A \cos 6 A+\cos 7 A} \\ &=\frac{\sin 7 A(2 \cos 6 A+1)}{\cos 7 A(2 \cos 6 A+1)} \\ &=\tan 7 A \end{aligned}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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