Question 714114
<pre>
I'll just do the first one:

tan(t) + {{{cos(t)/(1+sin(t))}}} = sec(t)

Rationalize the denominator of the second term by
multiplying it by {{{(conjugate_of_denominator)/(conjugate_of_denominator)}}}{{{""=""}}}{{{(1-sin(t))/(1-sin(t))}}}

tan(t) + {{{cos(t)/(1+sin(t))}}}{{{""*""}}}{{{(1-sin(t))/(1-sin(t))}}}

tan(t) + {{{(cos(t)(1-sin(t)))/((1+sin(t))(1-sin(t)))}}}

tan(t) + {{{(cos(t)(1-sin(t)))/(1-sin^2(t))}}}

You know the identity {{{sin^2(theta)+cos^2(theta)=1}}}.  When it
is solved like this {{{cos^2(theta)=1-sin^2(theta)}}} the right side
is the same form as the denominator and so we can write {{{1-sin^2(t)}}}
as {{{cos^2(t)}}}

tan(t) + {{{(cos(t)(1-sin(t)))/cos^2(t)}}}

And we cancel the cos(t) on top into the cosē(t) in the bottom

tan(t) + {{{(cross(cos(t))(1-sin(t)))/cos^cross(2)(t)}}}  

tan(t) + {{{(1-sin(t))/cos(t)}}}

Now we substitute {{{sin(t)/cos(t)}}} for {{{tan(t)}}}:

{{{sin(t)/cos(t)}}}{{{""+""}}}{{{(1-sin(t))/cos(t)}}}

Combine the fractions (they already have a common denominator):

{{{(sin(t)+1-sin(t))/cos(t)}}}

{{{(cross(sin(t))+1-cross(sin(t)))/cos(t)}}}

{{{1/cos(t)}}}

{{{sec(t)}}}

Edwin</pre>