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

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jgeertz | College Teacher | (Level 1) Associate Educator

Posted April 24, 2013 at 1:38 PM via web

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Prove this trig identity:

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

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted April 24, 2013 at 2:05 PM (Answer #1)

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`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.

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pramodpandey | College Teacher | Valedictorian

Posted April 24, 2013 at 2:11 PM (Answer #2)

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`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`

``

``

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oldnick | Valedictorian

Posted April 25, 2013 at 1:26 AM (Answer #3)

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`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`

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