Question 766479
{{{(csc (theta))(tan (theta))=se(theta)}}}

start with {{{(csc (theta))(tan (theta))}}}

{{{(csc (theta))(tan (theta))}}}......substitute {{{(tan (theta))=sec(theta)/(csc (theta))}}}

={{{(csc (theta))(sec(theta)/(csc (theta)))}}}


={{{cross((csc (theta)))1(sec(theta)/cross((csc (theta)))1)}}}


={{{sec(theta)}}}