Simplify

1. (5cosx/sin^{2}x)*[(sin^{2}x-sinxcosx)/(sin^{2}x-cos^{2}x)]
book answer is 5cotx/sinx+cosx

2.[1/(sin^{2}x-cos^{2}x)]-[2/(cosx-sinx)]
book answer is (1+2sinx+2cosx)/(sin

^{2}x-cos

^{2}x)

I've tried many manners of factoring/substitution

but can't seem to grasp these particular problems, please help.

On an unrelated note, how is it you are creating and pasting math images into this forum so easily?

i've tried using open office's math program and the filetype seems to be unacceptable(.odf).

$\displaystyle {5cos \over sin^2 x}*{{sin^2 x-sinxcosx} \over {sin^2 x-cos^2 x}}$

$\displaystyle {5cotx \over {sin x+cos x}}$

$\displaystyle {1 \over {sin^2 x-cos^2 x}}- {2 \over cos x -sin x}$

$\displaystyle {1+2sinx+2cosx} \over {sin^2 x-cos^2 x}$

Thank you very much sir!