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.
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