# Prove that `(1-cos^2x)(1+cot^2x)=1`

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When doing a problem like this, you only want to manipulate the left side. Make the left side equal to the right.

`(1-cos^2(x))(1-cot^2(x)) = 1`

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

Use foil

`1 - (cos^2(x))/(sin^2(x)) - cos^2(x) - (cos^4(x))/(sin^2(x)) = 1`

Get a common denominator and combine fractions (multiply the cos^2 by sin^2/sin^2)

`1 - (cos^2(x) - cos^2(x)*sin^2(x) - cos^4(x))/(sin^2(x)) = 1`

Factor out from the numerator

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

`1 - (cos^2(x) * 0 / (sin^2(x))) = 1`

`1 - 0 = 1`

Use the basic relations: `sin^(2)x + cos^(2) x=1 and cot^(2)x = [cos^(2)x]/[sin^(2)x]`

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Using these relations, we can make simplifications: (1-cos^(2)x) = sin^(2) x

and (1 + cot^(2)x) = (1 + cos^(2)x/sin^(2)x) = 1/sin^(2)x

Multiplying the two we get: sin^(2)x. 1/sin^(2)x = 1= right hand side. Hence proved.

cheers.

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This identity can be proven by applying Pythagorean identity, `sin^2x + cos^2x = 1` , several times.

First, we can rewrite the expression in first parenthesis as

`1-cos^2x = sin^2x` .

Also, recall that `cotx = cosx/sinx`

Then, `(1-cos^2x)(1+cot^2x) = sin^2x(1 + (cos^2x)/(sin^2x))`

Multiplying this out results in

`sin^2x + cos^2x = 1` , according to Pythagorean identity.

This means we have proven that

`(1-cos^2x)(1+cot^2x) = 1` , as required.

1-cos squared x=sin squared x

and cot x=sinX/cosX

so = sin squared x(1+cos squared X /sin squared X

so sin would multiply what is in the bracket and it would cancel sin in the bracket so=sin squared X+cos squared X

which is obviously=1