# proving identities

• Nov 5th 2008, 11:19 PM
john doe
proving identities
$
\frac{1+cot^3 t}{1+cot t}=csc^2 t - cot t
$

• Nov 5th 2008, 11:43 PM
mr fantastic
Quote:

Originally Posted by john doe
$
\frac{1+cot^3 t}{1+cot t}=csc^2 t - cot t
$

You can make life easier by factorising the sum of two cubes in the numerator of the right hand side expession and then cancelling the common factor.
• Nov 6th 2008, 12:02 AM
john doe
thns
• Nov 6th 2008, 05:55 AM
Soroban
Hello, john doe!

Mr. F is absolutely correct . . .

Quote:

$\frac{1+\cot^3\!t}{1+\cot t}\:=\:\csc^2\!t - \cot t$
Factor that sum-of-cubes . . .

$\text{We have: }\;\frac{(1+\cot t)(1 - \cot t + \cot^2\!t)}{1 + \cot t} \;\;=\;\;\underbrace{1 + \cot^2\!t}_{\text{This is }\csc^2\!t} - \cot t \;\;=\;\;\csc^2\!t - \cot t$