Trigonometric Identities (2)

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.(Clapping)

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).(Cool)

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

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

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

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

Thank you very much sir!

Re: Trigonometric Identities (2)

For eacj of these note that sin^2x - cos^2x is a difference of squares can be factored to (sinx-cosx)(sinx+cosx). That's really the only trick needed.

As or math notation - you cannot cuts and paste from other applications. Insead you can use LaTeX notation, placed between [tex ] and [/tex] tags. See the "LaTeX Help" section for info.