Question

(tgα+tgβ)/(cotgα+cotgβ)=tgα+tgβ

Answer 1

Respuesta:

α=180°

β=180°

Explicación paso a paso:

$\frac{ \tan( \alpha ) + \tan( \beta )}{ \cot( \alpha ) + \cot( \beta ) } = \tan( \alpha ) + \tan( \beta ) \\ 1 = \cot( \alpha ) + \cot( \beta ) \\ - \cot( \alpha ) = \cot( \beta ) \\ - \frac{1}{ \tan( \alpha ) } = \frac{1}{ \tan( \beta ) } \\ - \tan( \beta ) = \tan( \alpha )$

ó

$\frac{ \tan( \alpha ) + \tan( \beta )}{ \cot( \alpha ) + \cot( \beta ) } = \tan( \alpha ) + \tan( \beta ) \\ \frac{ \tan( \alpha ) + \tan( \beta ) }{ \frac{1}{ \tan( \alpha ) } + \frac{1}{ \tan( \beta ) } } = \tan( \alpha ) + \tan( \beta ) \\ \frac{ \tan( \alpha ) + \tan( \beta ) }{ \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) } } = \tan( \alpha ) + \tan( \beta ) \\ \tan( \alpha ) \tan( \beta ) = \tan( \alpha ) + \tan( \beta ) \\ \tan( \alpha ) \tan( \beta ) = - \tan( \beta ) + \tan( \beta ) \\ \tan( \alpha ) \tan( \beta ) = 0 \\ \tan( \alpha ) = 0; \: \tan( \beta ) = 0 \\ \alpha =180° ; \: \beta = 180°$