How do you prove cscx /(1 + cscx) + cscx /(1 - cscx) = (- 2sinx) / cos^2x?

1 Answer
May 6, 2015

I would start as:

(cscx(1-cscx)+cscx(1+cscx))/((1+cscx)(1-cscx))=-2sinx/cos^2x
(cscxcancel(-csc^2x)+cscxcancel(+csc^2x))/(1-csc^2x)=-2sinx/cos^2x
(2csc(x))/(1-csc^2x)=-2sinx/cos^2x

I would use the fact that:
cscx=1/sinx
So:
cancel(2)/cancel(sinx)(cancel(sin)^cancel(2)x)/(sin^2x-1)=-cancel(2)cancel(sinx)/cos^2x collect -1 on the left side:
-1/(1-sin^2x)=-1/cos^2x
But 1-sin^2x=cos^2x