• # question_answer Prove that: $\frac{\sin \,A-2\,{{\sin }^{3}}A}{2\,co{{s}^{3}}A-\cos \,A}=\tan \,A$.

 L.H.S. $=\frac{\sin \,A-2\,{{\sin }^{3}}A}{2\,{{\cos }^{3}}A-\cos \,A}$ $=\frac{\sin \,A\left( 1-2\,{{\sin }^{2}}A \right)}{\cos \,A\left( 2{{\cos }^{2}}\,A-1 \right)}$ $=\frac{\sin \,A}{\cos \,A}\frac{\left( 1-2\,{{\sin }^{2}}A \right)}{\left[ 2\left( 1-{{\sin }^{2}}A \right)-1 \right]}$ [$\because$ ${{\cos }^{2}}A=1-{{\sin }^{2}}A$] $=\frac{\sin \,A}{\cos \,A}\frac{(1-2\,si{{n}^{2}}A)}{(2-2\,si{{n}^{2}}A-1)}$ $=\frac{\sin \,A}{\cos \,A}\frac{(1-2\,si{{n}^{2}}A)}{(1-2\,si{{n}^{2}}A)}$ $=\tan \,A=\,R.H.S.$                          Hence Proved.