# Tan y - Sen y / Sen^3 y = Sec y / 1 + Cos y

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### 1 Answer

Verify the identity `(tany-siny)/(sin^3y)=(secy)/(1+cosy) `

We will show that the left side is equivalent to the right side:

`(tany-siny)/(sin^3y)=((siny)/(cosy)-siny)/(sin^3y) `

Multiply numerator and denominator by cosy:

`=(siny-sinycosy)/(cosysin^3y) `

Factor out the common siny in the numerator and divide to get:

`(1-cosy)/(cosysin^2y) `

Replace the sin^2y using the Pythagorean identity:

`=(1-cosy)/(cosy(1-cos^2y)) `

Factor the denominator:

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

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

`=1/(cosy(1+cosy)) `

`=1/(cosy)*1/(1+cosy) `

`=(secy)/(1+cosy) ` as required.

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