Question

tanA+1/b tan A=secA cosecA​

Answer 1

Answer:

$\tan( \alpha ) + \frac{1}{ \tan( \alpha ) } \\ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{1}{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } \\ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ \frac{ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} }{ \sin( \alpha ). \cos( \alpha ) } \\ \frac{1}{ \sin( \alpha ) \cos( \alpha ) } \\ \sec( \alpha ) . \csc( \alpha )$

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