Question

Tana-tanb/cotb-cota=tana*tanb

Answer 1

cot is an inverse function or rival of tan:

`{ \boxed{ \bf{ \cot( \theta) = (1)/( \tan( \theta) ) }}}`

Considering the question:

`{ \tt{ ( \tan( a) - \tan(b) )/( \cot(b) - \cot(a) ) = \tan(a) . \tan(b) }} \\ \\ { \tt{ \tan(a) - \tan(b) = ( \tan(a). \tan(b) )( \cot(b) - \cot(a) ) }} \\ { \tt{ \tan(a) - \tan(b) = \tan(a) \cot(b) \tan(b) - \cot(a) \tan(a) \tan(b) }} \\ \\ { \tt{ \tan(a) - \tan(b) = ( \tan(a) \tan(b) )/( \tan(b) ) - ( \tan(a) \tan(b) )/( \tan(a) ) }} \\ \\ { \tt{ \tan(a) - \tan(b) = \tan(a) - \tan(b) }}`

#Hence L.H.S = R.H.S, equation is consistent.