Question 376402
{{{(cot^3(x) - tan^3(x))/(sec^2(x) + cot^2(x)) = 2cot(2x)}}}
First, we can use the pattern, {{{a^3-b^3 = (a-b)(a^2+ab+b^2)}}}, to factor the numerator:
{{{((cot(x) - tan(x))(cot^2(x) + cot(x)tan(x) + tan^2(x)))/(sec^2(x) + cot^2(x)) = 2cot(2x)}}}
Since cot and tan are reciprocals, cot(x)*tan(x) = 1:
{{{((cot(x) - tan(x))(cot^2(x) + 1 + tan^2(x)))/(sec^2(x) + cot^2(x)) = 2cot(2x)}}}
Since {{{1 + tas^2(x) = sec^2(x)}}}:
{{{((cot(x) - tan(x))(cot^2(x) + sec^2(x)))/(sec^2(x) + cot^2(x)) = 2cot(2x)}}}
We can now see that the fraction will reduce:
{{{((cot(x) - tan(x))cross((cot^2(x) + sec^2(x))))/cross((sec^2(x) + cot^2(x))) = 2cot(2x)}}}
leaving:
{{{cot(x) - tan(x) = 2cot(2x)}}}
Rewriting the left side using sin and cos (a common technique use in these problems) we get:
{{{cos(x)/sin(x) - sin(x)/cos(x) = 2cot(2x)}}}
Since the right side has one term, we probably want one term on the left side, too. So we will add the fractions. Of course we need common denominators first:
{{{(cos(x)/sin(x))(cos(x)/cos(x)) - (sin(x)/cos(x))(sin(x)/sin(x)) = 2cot(2x)}}}
{{{cos^2(x)/(sin(x)cos(x)) - sin^2(x)/(sin(x)cos(x)) = 2cot(2x)}}}
Now we can subtract:
{{{(cos^2(x)-sin^2(x))/(sin(x)cos(x)) = 2cot(2x)}}}
The numerator exactly matches the formula for cos(2x). The denominator is close to sin(2x). sin(2x) = 2sin(x)cos(x). If we multiply both sides by 1/2 we get: {{{(1/2)sin(2x) = sin(x)cos(x)}}}. Substituting these into our fraction we get:
{{{cos(2x)/((1/2)sin(x)cos(x)) = 2cot(2x)}}}
We can get rid of the fraction in the denominator by multiplying the numerator and denominator by 2:
{{{(cos(2x)/((1/2)sin(2x)))(2/2) = 2cot(2x)}}}
giving:
{{{(2cos(2x))/sin(2x) = 2cot(2x)}}}
or
{{{2(cos(2x)/sin(2x)) = 2cot(2x)}}}
The fraction is cot(2x):
2cot(2x) = 2cot(2x) Done!