SOLUTION: hi could you please help me to verify this identity (cot^3 x - tan^3 x)/(sec^2x + cot^2x) = 2cot(2x)

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Question 376402: hi could you please help me to verify this identity

(cot^3 x - tan^3 x)/(sec^2x + cot^2x) = 2cot(2x)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!

First, we can use the pattern, a%5E3-b%5E3+=+%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29, to factor the numerator:

Since cot and tan are reciprocals, cot(x)*tan(x) = 1:

Since 1+%2B+tas%5E2%28x%29+=+sec%5E2%28x%29:

We can now see that the fraction will reduce:

leaving:
cot%28x%29+-+tan%28x%29+=+2cot%282x%29
Rewriting the left side using sin and cos (a common technique use in these problems) we get:
cos%28x%29%2Fsin%28x%29+-+sin%28x%29%2Fcos%28x%29+=+2cot%282x%29
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:


Now we can subtract:
%28cos%5E2%28x%29-sin%5E2%28x%29%29%2F%28sin%28x%29cos%28x%29%29+=+2cot%282x%29
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: %281%2F2%29sin%282x%29+=+sin%28x%29cos%28x%29. Substituting these into our fraction we get:
cos%282x%29%2F%28%281%2F2%29sin%28x%29cos%28x%29%29+=+2cot%282x%29
We can get rid of the fraction in the denominator by multiplying the numerator and denominator by 2:
%28cos%282x%29%2F%28%281%2F2%29sin%282x%29%29%29%282%2F2%29+=+2cot%282x%29
giving:
%282cos%282x%29%29%2Fsin%282x%29+=+2cot%282x%29
or
2%28cos%282x%29%2Fsin%282x%29%29+=+2cot%282x%29
The fraction is cot(2x):
2cot(2x) = 2cot(2x) Done!