Question

$\frac{(\mathrm{sin\theta }-2{\sin }^3\mathrm{\theta })}{(2{\cos }^3\mathrm{\theta }-\mathrm{cos\theta })}=?$
(a) tan θ
(b) cot θ
(c) sec θ
(d) cosec θ

Answer 1

(a) tan θ
$$\begin{gathered} \left(\frac{\text{sin}\theta -2{\text{sin}}^3\theta }{2{\text{cos}}^3\theta -\text{cos}\theta }\right) \\ \text{=}\frac{\text{sin}\theta (1-2{\text{sin}}^2\theta )}{\text{cos}\theta (2{\text{cos}}^2\theta -1)} \\ \text{=}\frac{\text{sin}\theta }{\text{cos}\theta }\text{×}\frac{\text{cos2}\theta }{\text{cos2}\theta }\left[\begin{array}{l}\because (1-2{\text{sin}}^2\theta )=\text{cos2}\theta \\ (2{\text{cos}}^2\theta -1)=\text{cos2}\theta \end{array}\right] \\ \text{=tan}\theta \end{gathered}$$