192,932 views
5 votes
5 votes
Tana-tanb/cotb-cota=tana*tanb

User Andrew Surzhynskyi
by
2.4k points

1 Answer

9 votes
9 votes

Answer:

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.

User Helix
by
2.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.