I am unsure what the right working is for this identity.

Simplify:

`(1-cosx)/(sin^2x)`

cos^2x +sin ^2x = 1

:. 1-cosx = sinx

leaving sinx/sin^2x

I need help with the next line please. thanks in advance. :)

### 3 Answers | Add Yours

`1-cos(x)!=sin(x)`

What I iamgine ,you want to write

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

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

={(1-cos(x))(1+cos(x))}/{sin^2(x).(1+cos(x))}

=(1-cos^2(x))/{sin62(x)(1+cos(x))}

=sin^2(x)/{sin^2(x)(1+cos(x))}

=1/(1+cos(x))

=RHS.

1-cos(x) is not equal to sin(x)

What I iamgine ,you want to write

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

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

={(1-cos(x))(1+cos(x))}/{sin^2(x).(1+cos(x))}

=(1-cos^2(x))/{sin^2(x)(1+cos(x))}

=sin^2(x)/{sin^2(x)(1+cos(x))}

=1/(1+cos(x))

=RHS.

Now I have you correct question

Simplify:

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

Here ,we will use following identity

`sin^2(x)+cos^2(x)=1`

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

`=(1-cos(x))(1+cos(x))`

Thus

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

`=1/(1+cos(x))`

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