# Prove this trig identity: `cot theta/(csc theta -1) =( csc theta +1)/cot theta`

*print*Print*list*Cite

`cot theta/(csc theta -1) =( csc theta +1)/cot theta`

To prove, consider the left side of the equation only. Multiply its numerator and denominator by the conjugate of `csc theta -1` .

`cot theta/(csc theta -1)*(csc theta+1)/(csc theta+1) =( csc theta +1)/cot theta`

`(cot theta(csc theta +1))/((csc theta -1)(csc theta +1))=( csc theta +1)/cot theta`

`(cot theta (csc theta +1))/(csc^2theta-1)=( csc theta +1)/cot theta`

Then, apply the Pythagorean identity `1 + cot^2A=csc^2A` . So, `csc^2theta-1` becomes `cot^2theta` .

`(cot theta (csc theta+1))/(cot^2 theta)=( csc theta +1)/cot theta`

And, cancel common factor.

`(csc theta +1)/(cot theta)= (csc theta + 1)/(cot theta) ` (True)

**Since left side can be express with the same expression at the right side, this proves that the given equation is an identity.**

`cot(theta)/(cosec(theta)-1)=(cosec(theta)+1)/cot(theta)`

we know

`cosec^2(theta)=cot^2(theta)+1`

LHS=`{cot(theta)/(cosec(theta)-1)}{(cosec(theta)+1)/(cosec(theta)+1)}`

`` =`(cot(theta)(cosec(theta)+1))/(cosec^2(theta)-1)`

`=(cot(theta)(cosec(theta)+1))/(cot^2(theta))`

`=(cosec(theta)+1)/(cot(theta))`

`=RHS`

``

``

`cot theta /(csc theta -1)=(csc theta +1)/cot theta`

It means, we have to prove:

`cot^2 theta= csc^2 theta -1`

Indeed:

`cot^2 theta= 1/(sin^2theta) -1=(1-sin^2 theta)/(sin^2theta)=(cos^2theta)/(sin^2theta)=cot^2theta`