$\frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x}$ is equal to -

cot 3 x

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cos 3 x

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tan 3 x

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None of these

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Solution

The correct option is A $cot 3 x$
$\frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x}$
$= \frac{(cos 4 x + cos 2 x) + cos 3 x}{(sin 4 x + sin 2 x) + sin 3 x}$
$= \frac{2 cos 3 x cos x + cos 3 x}{2 sin 3 x cos x + sin 3 x}$
$= \frac{cos 3 x [2 cos x + 1]}{sin 3 x [2 cos x + 1]}$
$= cot 3 x$

Suggest Corrections

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