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Question

Solve : 
$$ \dfrac{\sin 3A+2\sin 5A +\sin 7A}{\sin A+2\sin 3A +\sin 5A}=\dfrac{\sin 5A}{\sin 3A}$$


Solution

$$\dfrac{\sin 3a+2\sin 5a +\sin 7a}{\sin a+2\sin 3a +\sin 5a}$$

$$=\dfrac{(\sin 3a+\sin 5a)( \sin 5a+\sin 7a)}{(\sin a+\sin 3a)( \sin 3a+\sin 5a)}$$

$$=\dfrac{\left \{ 2\sin 4a\cos a \right \}+\left \{ 2\sin 6a \cos a \right \}}{\left \{  2 \sin 2a \cos a\right \}+\left \{ 2 \sin 4a \cos a \right \}}$$

$$=\dfrac{\sin 4a+\sin 6a}{\sin 2a+\sin 4a}$$

$$=\dfrac{2\sin 5a \cos a}{2 \sin 3a \cos a}$$

$$=\dfrac{\sin 5a}{\sin 3a}$$

$$\therefore \dfrac{\sin 3A+2\sin 5A +\sin 7A}{\sin A+2\sin 3A +\sin 5A}=\dfrac{\sin 5A}{\sin 3A}$$

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