question_answer

Prove that: \[\frac{\sin \,A-2\,{{\sin }^{3}}A}{2\,co{{s}^{3}}A-\cos \,A}=\tan \,A\].

Answer:

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.