Question

The value of $\frac{\cos 3x}{2\cos 2x-1}$ is equal to
(a) cos x
(b) sin x
(c) tan x
(d) none of these

Answer 1

(a) cos x

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \therefore \frac{\cos 3x}{2\cos 2x-1}=\frac{4{\cos }^{3}x-3\cos x}{2\left(2{\cos }^{2}x-1\right)-1}\left[\because \cos 3x=4{\cos }^{3}x-3\cos x\right] \\ =\frac{4{\cos }^{3}x-3\cos x}{4{\cos }^{2}x-2-1} \\ =\frac{4{\cos }^{3}x-3\cos x}{4{\cos }^{2}x-3} \\ =\cos x\left(\frac{4{\cos }^{2}x-3}{4{\cos }^{2}x-3}\right) \\ =\cos x \end{gathered}$$