Prove that, (1+tan theta)(sin theta-cos theta)=(sec …

LHS = (1 + tan{tex}\theta{/tex} + cot{tex}\theta{/tex})(sin{tex}\theta{/tex} - cos{tex}\theta{/tex})
{tex}= \left( 1 + \frac { \sin \theta } { \cos \theta } + \frac { \cos \theta } { \sin \theta } \right) ( \sin \theta - \cos \theta ){/tex}
{tex}= \left( \frac { \cos \theta \sin \theta + \sin ^ { 2 } \theta + \cos ^ { 2 } \theta } { \cos \theta \sin \theta } \right) ( \sin \theta - \cos \theta ){/tex}
{tex}= \frac { ( \cos \theta \sin \theta + 1 ) } { \cos \theta \sin \theta } ( \sin \theta - \cos \theta ){/tex}
RHS = {tex}\left( \frac { \sec \theta } { \ cosec ^ { 2 } \theta } - \frac { \cos e c \theta } { \sec ^ { 2 } \theta } \right) = \left( \frac { \frac { 1 } { \cos \theta } } { \frac { 1 } { \sin ^ { 2 } \theta } } - \frac { \frac { 1 } { \sin \theta } } { \frac { 1 } { \cos ^ { 2 } \theta } } \right){/tex}
{tex}= \left( \frac { \sin ^ { 2 } \theta } { \cos \theta } - \frac { \cos ^ { 2 } \theta } { \sin \theta } \right) = \frac { \sin ^ { 3 } \theta - \cos ^ { 3 } \theta } { \cos \theta \sin \theta }{/tex}
{tex}= \frac { ( \sin \theta - \cos \theta ) \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta + \cos \theta \sin \theta \right) } { \cos \theta \sin \theta }{/tex} [{tex}\because{/tex} a³ - b³ = (a - b) (a² + ab + b²) ]
{tex}= \frac { ( \sin \theta - \cos \theta ) ( 1 + \cos \theta \sin \theta ) } { \cos \theta \sin \theta }{/tex}
{tex}\therefore{/tex} LHS = RHS