Question 148633
Prove the identity:

{{{(Csc^2x-Cot^2x)/(Sec^2x)=Cos^2x}}}
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Make two fractions out of the left side:

{{{(Csc^2x)/(Sec^2x)-(Cot^2x)/(Sec^2x)}}}

Substitute {{{1/(Sin^2x)}}} for {{{Csc^2x}}}.
Substitute {{{1/(Cos^2x)}}} for both {{{Sec^2x}}}'s.
Substitute {{{(Cos^2x)/(Sin^2x)}}} for {{{Cot^2x}}}.

{{{(1/(Sin^2x))/(1/(Cos^2x))-((Cos^2x)/(Sin^2x))/(1/(Cos^2x))}}}

Invert and multiply:

{{{(1/(Sin^2x))((Cos^2x)/1)-((Cos^2x)/(Sin^2x))((Cos^2x)/1)}}}

Simplify:

{{{(Cos^2x)/(Sin^2x)-(Cos^4x)/(Sin^2x)}}}

Combine the numerators over the common denominator:

{{{(Cos^2x-Cos^4x)/(Sin^2x)}}}

Factor {{{Cos^2x}}} out of the numerator:

{{{(Cos^2x(1-Cos^2x))/(Sin^2x)}}}

Replace {{{1-Cos^2x}}} by {{{Sin^2x}}}

{{{(Cos^2x(Sin^2x))/(Sin^2x)}}}

Cancel the {{{Sin^2x}}}'s

{{{(Cos^2x(cross(Sin^2x)))/(cross(Sin^2x))}}}

{{{Cos^2x}}}

Edwin</pre>