# The value of

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,

$\therefore \frac{\cos 3 x}{2 \cos 2 x-1}=\frac{4 \cos ^{3} x-3 \cos x}{2\left(2 \cos ^{2} x-1\right)-1} \quad\left[\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right]$

$=\frac{4 \cos ^{3} x-3 \cos x}{4 \cos ^{2} x-2-1}$

$=\frac{4 \cos ^{3} x-3 \cos x}{4 \cos ^{2} x-3}$

$=\cos x\left(\frac{4 \cos ^{2} x-3}{4 \cos ^{2} x-3}\right)$

$=\cos x$