[sec(A) + csc(A)] ÷ [1+ tan(A)]= csc(A)

Recall the following trig definitions:

sec(A) = 1 / cos(A);
csc(A) = 1 / sin(A);
tan(A) = [sin(A) / cos(A)];
1 = [sin(A) / sin(A)]

Substituting the definitions of sec(A), csc(A), and tan(A), in terms of sin(A)
and cos(A), we have the following:

[sec(A) + csc(A)] / [ 1 + tan(A)] =

{ [1 / cos(A)] + [1 / sin(A)] } / { [sin(A) / sin(A)] + [sin(A) / cos(A)] }
The right side of the above equation becomes:

{ [ sin(A) + cos(A) ] / sin(A)cos(A) } / { sin(A)cos(A) + [sin(A)]^2 / sin(A)cos(A) }

Inverting and multiplying the denominator of the above equation, results in:

{ [ sin(A) + cos(A) ] / [sin(A)cos(A)] } x { sin(A)cos(A) / [sin(A)] ^ 2 + sin(A)cos(A)}

The term sin(A)cos(A) cancels out of the above equation, and we have:

[sin(A) + cos(A)] / { [sin(A)] ^ 2 + [sin(A)cos(A)] }

Factoring sin(A) out of the denominator of the above equation, we have:

[sin(A) + cos(A)] / { sin(A) [ sin(A) + cos(A)] }

The sum sin(A) + cos(A) cancels out of the numerator and the denominator,
resulting in:

[ 1 / sin (A)] = csc(A) (Which was to be demonstrated).

Thanks for your help. JackieF