Question 979423
<pre>
{{{ (tan^3(x) - 1)/ (tan(x) - 1) - sec^2(x) + 1 = 0 }}}

Factor the numerator as the difference of cubes: {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}

Since the denominator tan(x)-1 cannot be 0, tan(x) cannot be 1, so x
cannot be {{{pi/4+2pi*n}}} or {{{5pi/4+2pi*n}}}

{{{ (cross((tan(x) - 1))(tan^2(x)+tan(x)+1))/ (cross(tan(x) - 1)) - sec^2 (x) + 1 = 0 }}}

{{{ tan^2(x) +tan(x) + 1 - sec^2(x) + 1 = 0 }}}

{{{ tan^2(x) +tan(x) - sec^2(x) + 2 = 0 }}}

{{{ tan^2(x) +tan(x) - sec^2(x) + 2 = 0 }}}

Now we use the identity {{{1+tan^2(theta)=sec^2(theta)}}} solved for
                        {{{tan^2(theta)=sec^2(theta)-1}}}

{{{ (sec^2(x)-1) +tan(x) - sec^2(x) + 2 = 0 }}}

{{{ sec^2(x)-1 +tan(x) - sec^2(x) + 2 = 0 }}}

{{{tan(x)+1=0}}}

{{{tan(x)=-1}}}

{{{x= 3pi/4+2pi*n}}} or {{{x = 7pi/4+2pi*n}}}

{{{x= 3pi/4+8pi*n/4}}} or {{{x = 7pi/4+8pi*n/4}}}

{{{x= (3pi+8pi*n)/4}}} or {{{x = (7pi+8pi*n)/4}}}

{{{x= pi(3+8n)/4}}} or {{{x = pi(7+8n)/4}}}

Edwin</pre>