Question 635241
<pre>
{{{1/(1+cos(x))}}} + {{{cos(x)/(1-cos(x))}}}

Since csc(x) = {{{1/sin(x)}}} we'll get it first in terms of sin(x)

Get an LCD of {{{(1 + cos(x))(1-cos(x))}}}

{{{1/(1+cos(x))}}}·{{{(1 - cos(x))/(1-cos(x))}}} + {{{cos(x)/(1-cos(x))}}}·{{{(1 + cos(x))/(1+cos(x))}}}



{{{(1 - cos(x))/(1-cos^2(x))}}} + {{{(cos(x)(1 + cos(x)))/(1-cos^2(x))}}}

Since the denominators 1-cosē(x) are equal to sinē(x)

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

Multiply the numerator of the second fraction out:

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

Combine the numerators over the common denominator

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

{{{(1+ cos^2(x))/sin^2(x)}}}

Replace cosē(x) by 1-sinē(x)

{{{(1+ (1-sin^2(x)))/sin^2(x)}}}

{{{(1+ 1-sin^2(x))/sin^2(x)}}}

{{{(2-sin^2(x))/sin^2(x)}}}

Break into two fractions:

{{{2/sin^2(x)}}} - {{{sin^2(x)/sin^2(x)}}}

{{{2/sin^2(x)}}} - {{{cross(sin^2(x))/cross(sin^2(x))}}}

{{{2/sin^2(x)}}} - 1

Bring sinē(x) to the numerator as cscē(x)

2cscē(x) - 1

Edwin</pre>