Question 732084
<pre>
{{{(sin^2(theta)-tan(theta))/(cos^2(theta)-cot(theta))}}} = tan²(<font face="symbol">q</font>)

In the left side:

Replace tan(<font face="symbol">q</font>) with {{{sin(theta)/cos(theta)}}}
and
Replace cot(<font face="symbol">q</font>) with {{{cos(theta)/sin(theta)}}}

{{{(sin^2(theta)-sin(theta)/cos(theta))/(cos^2(theta)-cos(theta)/sin(theta))}}}

Write as a division:

{{{(sin^2(theta)-sin(theta)/cos(theta))}}}{{{"÷"}}}{{{(cos^2(theta)-cos(theta)/sin(theta))}}}

Get a common denominator in each:

{{{(sin^2(theta)cos(theta)/cos(theta)-sin(theta)/cos(theta))}}}{{{"÷"}}}{{{(cos^2(theta)sin(theta)/sin(theta)-cos(theta)/sin(theta))}}}

{{{(sin^2(theta)cos(theta)-sin(theta))/cos(theta)}}}{{{"÷"}}}{{{(cos^2(theta)sin(theta)-cos(theta))/sin(theta)}}}

Invert the second fraction and change division to multiplication:

{{{(sin^2(theta)cos(theta)-sin(theta))/cos(theta)}}}{{{""*""}}}{{{sin(theta)/(cos^2(theta)sin(theta)-cos(theta))}}}

Factor sin(<font face="symbol">q</font>)out of left numerator
and factor cos(<font face="symbol">q</font>) out of right denominator:

{{{sin(theta)(sin(theta)cos(theta)-1)/cos(theta)}}}{{{""*""}}}{{{sin(theta)/
(cos(theta)(cos(theta)sin(theta)-1))}}}

The two expressions in parentheses will cancel:

{{{sin(theta)(cross(sin(theta)cos(theta)-1))/cos(theta)}}}{{{""*""}}}{{{sin(theta)/
(cos(theta)(cross(cos(theta)sin(theta)-1)))}}}

{{{(sin(theta)*sin(theta))/(cos(theta)*cos(theta))}}}

{{{sin^2(theta)/cos^2(theta)}}}

{{{(sin(theta)/cos(theta))^2}}}

tan²(<font face="symbol">q</font>)

Edwin</pre>