Question

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

Answer 1

(a) $\cos \mathrm{\theta }$

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