Question

Verify cotB+tanB=secBcscB

Answer 1

Hmm let me stab at this
has. been a while and identifies aren't fresh.
$\frac{1}{ \tan( \alpha ) } + \tan( \alpha ) \\ = \frac{1}{ \tan( \alpha ) } + \frac{ { \tan( \alpha ) }^{2} }{ \tan( \alpha ) } \\ = \frac{1 + { \tan( \alpha ) }^{2} }{ \tan( \alpha ) } \\ = \frac{ { \sec( \alpha ) }^{2} }{ \tan( \alpha ) }$
so then we get
$\sec( \alpha ) ( \frac{ \sec( \alpha ) }{ \tan( \alpha ) } ) \\ = \sec( \alpha ) ( \frac{ \frac{1}{ \cos( \alpha )} }{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } ) \\ = \sec( \alpha ) ( \frac{1}{ \sin( \alpha ) } ) \\ = \sec( \alpha ) \csc( \alpha )$
vote brainliest if you like my answer!