(A) Prove that (sin A -2 sin^3 A)/(2 cos^3 A - cos A)= tan A

\[\begin{aligned} L H S= & \frac{\sin A}{} 2 \cos ^{3} A-\cos A \\ 2 \cos ^{3} A & =\frac{\sin A\left(1-2 \sin ^{2} A\right)}{\cos A\left(2 \cos ^{2} A-1\right)} \\ & =\frac{\sin A\left[1-2\left(1-\cos ^{2} A\right)\right]}{\cos A\left[2 \cos ^{2} A-1\right]}=\frac{\sin A\left[1-2+2 \cos ^{2} A\right]}{\cos A\left[2 \cos ^{2} A-1\right]} \\ & =\frac{\sin A\left[2 \cos ^{2} A-1\right]}{\cos A\left[2 \cos ^{2} A-1\right]}=\tan A=\text { RHS } \end{aligned} \]

\[\begin{aligned} \mathrm{LHS}= & \sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A}) \\ & =\frac{1}{\cos \mathrm{~A}}(1-\sin \mathrm{A})\left(\frac{1}{\cos \mathrm{~A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{~A}}\right) \\ & =\frac{1}{\cos \mathrm{~A}}(1-\sin \mathrm{A}) \frac{(1+\sin \mathrm{A})}{\cos \mathrm{A}} \\ & =\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=1=\mathrm{RHS} \end{aligned} \]