# I understand everything up to step 5. But I fail to see how they got to step 6. I get an answer of sinx + 1 / sin^2x-1 ........

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`(1/(sinx+1))-(2/(cos^2 x))`

` ` `(cos^2 x - 2(sinx+1))/((sinx+1)cos^2 x)`

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

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

`-((sin^2 x + 2sinx + 1))/((sinx+1)(1-sin^2 x))`

`-((sinx+1)(sinx+1))/((sinx+1)(1-sinx)(1+sinx))`

` ` `(-(sinx+1))/(-(sinx-1)(sinx+1)`

` `

`1/(sinx - 1)`

So the answer can be either of the last two steps. There is error in the identity. You have arrived at the correct answer.

` `

No the answer is **false which you attached** .

Lets solve and see what we get

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

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

= `(1/(sin x +1))- (2/((1-sin x)*(1+sin x)))`

= `(1-sinx - 2)/((1-sin x)*(1+sin x))`

= `(-1 -sinx)/((1-sin x)*(1+sin x))`

= `(-(1+sin x))/((1-sin x)*(1+sin x))`

= `(- 1)/(1-sin x)`

=` 1/(sin x -1)` is the answer

But as i see the pick you have uploaded , i think there is a typing mistake .In the R.H.S if the denominator is sin^2 x -1 instead of sin x - 1 , then what you have provided is true.

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

=`1/(sin x -1) ` [ as proved above]

= `(1/(sin x -1))*((sin x+1)/ (sin x+1))`

= `(sin x+1)/((sinx-1)(sinx+1))`

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

so , i think its a typing mistake