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Question
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
Options
cos x
sin x
tan x
none of these
Solution
cos x
\[\text{ We have } , \]
\[ \therefore \frac{\cos3x}{2\cos2x - 1} = \frac{4 \cos^3 x - 3\text{ cos } x}{2\left( 2 \cos^2 x - 1 \right) - 1} \left[ \because \cos3x = 4 \cos^3 x - 3\text{ cos } x \right]\]
\[ = \frac{4 \cos^3 x - 3\text{cos } x}{4 \cos^2 x - 2 - 1}\]
\[ = \frac{4 \cos^3 x - 3\text{ cos } x}{4 \cos^2 x - 3}\]
\[ = \text{ cos } x\left( \frac{4 \cos^2 x - 3}{4 \cos^2 x - 3} \right) \]
\[ = \text{ cos } x\]
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