this problem is pretty hard !!
2007-11-16 11:03:01 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
cscx-sinx = 1/sinx - sinx = (1-sin^2x)/sin
= cos^2 x / sinx
= cosx/(1/cosx * sinx)
= cosx/(secx * sinx)
2007-11-16 11:10:34 · answer #1 · answered by norman 7 · 2⤊ 0⤋
cotx=cosx/sinx and cosecx=a million/ sinx employing the two identities L.H.S cosx/sinx - a million/ sinx (cosx+a million) taking LCM i.e sinx (cosx - a million/ sinx )(cosx+a million) multiplying the numerators cos^2x+cosx-cosx-a million / sinx cos^2x-a million/ sinx by way of fact the identity is cos^2x+ sin^2x=a million - sin^2x/ sinx - sinx--------------------that's equals acceptable hand area , shown
2016-11-11 20:51:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
i see sex...i can verify what that is!
2007-11-16 11:10:38 · answer #3 · answered by matt 3 · 0⤊ 1⤋