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Alika [10]
3 years ago
15

Tana-tanb/cotb-cota=tana*tanb

Mathematics
1 answer:
Tems11 [23]3 years ago
6 0

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.

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