Question:
The value of $\frac{\cos 3 x}{2 \cos 2 x-1}$ is equal to
(a) cos x
(b) sin x
(c) tan x
(d) none of these
Solution:
(a) cos x
We have,
$\therefore \frac{\cos 3 x}{2 \cos 2 x-1}=\frac{4 \cos ^{3} x-3 \cos x}{2\left(2 \cos ^{2} x-1\right)-1} \quad\left[\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right]$
$=\frac{4 \cos ^{3} x-3 \cos x}{4 \cos ^{2} x-2-1}$
$=\frac{4 \cos ^{3} x-3 \cos x}{4 \cos ^{2} x-3}$
$=\cos x\left(\frac{4 \cos ^{2} x-3}{4 \cos ^{2} x-3}\right)$
$=\cos x$
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