mendez1369p7zc2z
mendez1369p7zc2z
30-04-2018
Mathematics
contestada
verify cotB+tanB=secBcscB
Respuesta :
sheavner
sheavner
30-04-2018
hmm let me stab at this
has. been a while and identifies aren't fresh.
[tex] \frac{1}{ \tan( \alpha ) } + \tan( \alpha ) \\ = \frac{1}{ \tan( \alpha ) } + \frac{ { \tan( \alpha ) }^{2} }{ \tan( \alpha ) } \\ = \frac{1 + { \tan( \alpha ) }^{2} }{ \tan( \alpha ) } \\ = \frac{ { \sec( \alpha ) }^{2} }{ \tan( \alpha ) }[/tex]
so then we get
[tex] \sec( \alpha ) ( \frac{ \sec( \alpha ) }{ \tan( \alpha ) } ) \\ = \sec( \alpha ) ( \frac{ \frac{1}{ \cos( \alpha )} }{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } ) \\ = \sec( \alpha ) ( \frac{1}{ \sin( \alpha ) } ) \\ = \sec( \alpha ) \csc( \alpha ) [/tex]
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