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Question

Prove: $$\cfrac { \sin A+\sin 3A+\sin 5A+\sin 7A }{ \cos A+\cos 3A+\cos 5A+\cos 7A }=\tan x$$
then $$x$$ is equal to,


A
4A
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B
3A
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C
2A
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D
A
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Solution

The correct option is A 4A
$$\cfrac { \sin A+\sin 3A+\sin 5A+\sin 7A }{ \cos A+\cos 3A+\cos 5A+\cos 7A }=\tan x$$

$$=\cfrac { (\sin A+\sin 7A)+(\sin 3A+\sin 5A) }{ (\cos A+\cos 7A)+(\cos 3A+\cos 5A) } $$

$$=\cfrac { 2\sin 4A\cos 3A+2\sin 4A\cos A}{ 2\cos 4A\cos 3A+2\cos 4A+\cos A } $$

$$=\cfrac { 2\sin 4A(\cos 3A+\cos A)}{ 2\cos 4A(\cos 3A+\cos A) } $$

$$=\cfrac{\sin 4A}{\cos 4A}$$

$$=\tan 4A=\tan x$$

$$x=4A$$

Maths

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