Question 433555
{{{Sec^4x - Sec^2x = Tan^4x + Tan^2x}}}
<pre>
To do this one we need the identity 
{{{1+Tan^2theta=Sec^2theta}}} and its rearranged version {{{Sec^2theta-1=Tan^2theta}}}

{{{Sec^4x - Sec^2x = Tan^4x + Tan^2x}}}

Factor common factor {{{Sec^2x}}} out of the left side

{{{Sec^2x(Sec^2x-1) = Tan^4x + Tan^2x}}}

Replace the first factor on the left, {{{Sec^2x}}}, by {{{1+Tan^x}}}

Replace the second factor on the left, {{{(Sec^2-1)}}} by {{{Tan^2x}}}


{{{(1+Tan^2x)(Tan^2x) = Tan^4x + Tan^2x}}}

Multiply the left side out:

{{{Tan^2x + Tan^4x =Tan^4x + Tan^2x}}}

Reverse the terms on the left

{{{Tan^4x + Tan^2x =Tan^4x + Tan^2x}}}

Edwin</pre>