Question 969538
NOTE:
I believe you meant {{{(1-sin(x))/cos(x)=(1-sin(x))/cos(x))}}} .
If that was not what you meant, then I solved the wrong problem.
 
You could write {{{(1-sin(x))/cos(x)=(1-sin(x))/cos(x))}}} as
(1-sinx)/cosx=cosx/(1+sinx) , because
1-sin(x)/cos(x)={{{1-sin(x)/cos(x)}}} and cos(x)/1+sin(x)={{{cos(x)/1+sin(x)}}} .
The parentheses are implicit in {{{(1-sin(x))/cos(x)=(1-sin(x))/cos(x))}}} .
The long horizontal line between numerator and denominator is a powerful grouping symbol that indicates that numerator and denominator must be calculated before doing the division.
When you cannot write two lines (as when you are entering calculations into a calculator or spreadsheet) you need to write the implicit parentheses.
 
The expressions in {{{(1-sin(x))/cos(x)=cos(x)/(1+sin(x))}}} do not exist (are undefined) when the denominators are zero.
{{{cos(x)=0}}} happens for {{{x=pi/2}}} , {{{x=3pi/2}}} , and all co-terminal angles.
In general, we could say that {{{x=(2k+1)pi/2}}} for every integer {{{k}}}
makes {{{cos(x)=0}}} and cannot be a solution to the equation above.
 
The values of {{{x}}} that make {{{1+sin(x)=0}}}<--->{{{sin(x)=-1}}} are
{{{3pi/2}}} and all co-terminal angles.
Those values will be excluded if we exclude {{{x=(2k+1)pi/2}}} for every integer {{{k}}}
to make {{{cos(x)<>0}}} .
As long as {{{x<>(2k+1)pi/2}}} for every integer {{{k}}} ,
which makes {{{cos(x)<>0}}} and {{{cos(x)<>0}}} , we can multiply both sides of
{{{(1-sin(x))/cos(x)=cos(x)/(1+sin(x))}}} times {{{cos(x)(1+sin(x))}}}
to get the equivalent equations
{{{(1-sin(x))(1+sin(x))cos(x)/cos(x)=cos(x)cos(x)(1+sin(x))/(1+sin(x))}}}--->{{{(1-sin(x))(1+sin(x))=cos(x)cos(x)}}}--->{{{1-sin^2(x)=cos^2(x)}}} ,
and since the last equation is a trigonometric identity,
true for all values of {{{x}}} ,
the solution to {{{1-sin^2(x)=cos^2(x)}}} is all values of {{{x}}} ,
and the solution to {{{(1-sin(x))/cos(x)=cos(x)/(1+sin(x))}}}
is all values of {{{x}}} such that {{{x<>(2k+1)pi/2}}} for every integer {{{k}}} .