Question

Tana-tanb/cotb-cota=tana*tanb

Answer 1

Answer:

cot is an inverse function or rival of tan:

${ \boxed{ \bf{ \cot( \theta) = \frac{1}{ \tan( \theta) } }}}$

Considering the question:

${ \tt{ \frac{ \tan( a) - \tan(b) }{ \cot(b) - \cot(a) } = \tan(a) . \tan(b) }} \\ \\ { \tt{ \tan(a) - \tan(b) = ( \tan(a). \tan(b) )( \cot(b) - \cot(a) ) }} \\ { \tt{ \tan(a) - \tan(b) = \tan(a) \cot(b) \tan(b) - \cot(a) \tan(a) \tan(b) }} \\ \\ { \tt{ \tan(a) - \tan(b) = \frac{ \tan(a) \tan(b) }{ \tan(b) } - \frac{ \tan(a) \tan(b) }{ \tan(a) } }} \\ \\ { \tt{ \tan(a) - \tan(b) = \tan(a) - \tan(b) }}$

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