Question

Prove that $\frac{(\sin A-2{\sin }^{3}A)}{(2{\cos }^{3}A-\cos A)}=\tan A.$

Answer 1

$$\begin{gathered} \text{LHS=}\frac{(\sin A-2{\sin }^{3}A)}{(2{\cos }^{3}A-\cos A)} \\ \text{=}\frac{\sin A(1-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-1)} \\ \text{=}\tan A\left\{\frac{({\sin }^{2}A+{\cos }^{2}A-2{\sin }^{2}A)}{2{\cos }^{2}A-{\sin }^{2}A-{\cos }^{2}A}\right\}[\because {\sin }^{2}A+{\cos }^{2}A=1] \\ \text{=}\tan A\left\{\frac{({\cos }^{2}A-{\sin }^{2}A)}{({\cos }^{2}A-{\sin }^{2}A)}\right\} \\ \text{=}\tan A \\ \text{=RHS} \end{gathered}$$