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embizze | High School Teacher | (Level 2) Educator Emeritus

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Verify the identity `(cot^2x)/(cscx-1)=cscx+sin^2x+cos^2x `

We will work from the left side, and show that it is equivalent to the right side:

`(cot^2x)/(cscx-1)=((cos^2x)/(sin^2x))/(1/(sinx)-1) `

Multiply numerator and denominator by sin^2x:

`=(cos^2x)/(sinx-sin^2x) `

Use the Pythagorean identity to rewrite the numerator:

`=(1-sin^2x)/(sinx(1-sinx)) `

Factor the numerator:

`=((1+sinx)(1-sinx))/(sinx(1-sinx)) `

The (1-sinx) terms divide to 1 leaving:

`=(1+sinx)/sinx `

`=cscx+1 `

Again using the Pythagorean identity we get:

`cscx+sin^2x+cos^2x ` as required.


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