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`
`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]`
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.
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