# Prove the identity: `(sin^2 alpha)/(cos alpha*(1-cos alpha)) = sec alpha + 1/sec alpha`

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### 1 Answer

Let the angle be `alpha` , such that:

`(sin^2 alpha)/(cos alpha*(1 - cos alpha)) = sec alpha + 1/sec alpha`

You should remember that`sec alpha = 1/cos alpha =gt 1/sec alpha = cos alpha.`

Write the new form of the identity:

`(sin^2 alpha)/(cos alpha*(1 - cos alpha)) = 1/cos alpha + cos alpha`

You need to bring the fractions from the right side to a common denominator:

`(sin^2 alpha)/(cos alpha*(1 - cos alpha)) = (1+cos alpha)/cos alpha`

Multiplying both sides by (1 - cos alpha) yields:

`(sin^2 alpha)/cos alpha= ((1 - cos alpha)(1 + cos alpha))/cos alpha`

The special product from the right side gives the difference of two squares:

`(sin^2 alpha)/cos alpha= (1 - cos^2 alpha)/cos alpha`

`` `1 - cos^2 alpha = sin^2 alpha`

Equating both sides yields:

`sin^2 alpha = sin^2 alpha`

**The last line proves that the identity is checked.**