$\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A} =$

tan 2 A

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cot 3 A

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tan 3 A

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tan 4 A

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Solution

The correct option is C $tan 3 A$
$\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A}$
$= \frac{(sin A + sin 5 A) + (sin 4 A + sin 2 A)}{(cos 5 A + cos A) + (cos 4 A + cos 2 A)}$
$= \frac{2 sin 3 A cos 2 A + 2 sin 3 A cos A}{2 cos 3 A cos 2 A + 2 cos 3 A cos A}$
$= \frac{2 sin 3 A (cos 2 A + cos A)}{2 cos 3 A (cos 2 A + cos A)}$
$= tan 3 A$

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Similar questions

\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A} =

\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A} =

{cos}^{4} A - {cos}^{2} A = {sin}^{4} A - {sin}^{2} A

c o s^{4} A - c o s^{2} A = s i n^{4} A - s i n^{2} A

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