# How do I verify the identity `(cot^2 x -1) /(1+cot^2 x) = 1-2sin^2 x` ``

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

We know that: `cotx = cosx/sinx`

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

`Now we will use common denominator.`

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

Now we will reduce `sin^2 x ` and substitute `sin^2 x + cos^2 x = 1`

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

Now we know that `sin^2 x + cos^2 x = 1 ==> cos^2 x = 1- sin^2 x.`

`==> (1-sin^2 x) - sin^2 x = 1- 2sin^2 x`

`==> 1- 2sin^2 x = 1- 2sin^2 x`

``

Nevermind. I figured it out:

((cos^2)/(sin^2))-1)/((cos^2)/(sin^2))+1)

(((cos^2)-(sin^2))/(sin^2))/(((cos^2)+(sin^2))/(sin^2))

(cos^2)-(sin^2)

1-sin^2-sin^2

1-2sin^2 = 1-2sin^2

cot^2x-1/1+cot^2x=(1-2 sin^2x)

we know that cot=cos/sin

(cos^2x/sin^2x)-1/1+cos^2x/sin^2x

cos^2x-sin^2x/sin^2x/sin^2x+cos^2x/sin^2x (cancelling sin^2x)

(as we know from the identity sin^2x+Cos^2x=1)

cos^2x-sin^2x/1

1-sin^2x-sin^2x

=1-2sin^2x RHS hence proved