Question

How do you prove (cotx + cscx / sinx + tanx) = (cotx)(cscx)?

Answer 1

Verified below

Explanation:

`(cotx + cscx) / (sinx + tanx) = (cotx)(cscx)`

`(cosx/sinx + 1/sinx) / (sinx + sinx/cosx) = (cotx)(cscx)`

`((cosx + 1)/sinx) / ((sinxcosx)/cosx + sinx/cosx) = (cotx)(cscx)`

`((cosx + 1)/sinx) / ((sinx(cosx+1))/cosx) = (cotx)(cscx)`

`(cancel(cosx + 1)/sinx) * (cosx/(sinxcancel((cosx+1)))) = (cotx)(cscx)`

`(cosx/sinx*1/sinx) = (cotx)(cscx)`

`(cotx)(cscx) = (cotx)(cscx)`

Answer 2

We're trying to prove that `(cotx+cscx)/(sinx+tanx)=cotxcscx`. Here are the identities you'll need:

`tanx=sinx/cosx`

`cotx=cosx/sinx`

`cscx=1/sinx`

I'll start with the left side and manipulate it until it equals the right side:

`color(white)=(cotx+cscx)/(sinx+tanx)`

`=(qquadcosx/sinx+1/sinxqquad)/(qquadsinx/1+sinx/cosxqquad)`

`=(qquad(cosx+1)/sinxqquad)/(qquad (sinxcosx)/cosx+sinx/cosxqquad)`

`=(qquad(cosx+1)/sinxqquad)/(qquad (sinxcosx+sinx)/cosxqquad)`

`=(cosx+1)/sinx*cosx/(sinxcosx+sinx)`

`=(cosx+1)/sinx*cosx/(sinx(cosx+1))`

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

`=(cosxcolor(red)cancelcolor(black)((cosx+1)))/(sin^2xcolor(red)cancelcolor(black)((cosx+1)))`

`=cosx/sin^2x`

`=cosx/sinx*1/sinx`

`=cotx*cscx`

`=RHS`

That's the proof. Hope this helped!

Answer 3

`LHS=(cotx+cscx)/(sinx+tanx)`

`=(cotx+cscx)/(sinx+tanx)*((cotx*cscx)/(cotx*cscx))`

`=cotx*cscx[(cotx+cscx)/((sinx+tanx)*cotx*cscx)]`

`=cotx*cscx[(cotx+cscx)/((sinx*cscx*cotx+tanx*cotx*cscx))]`

`=cotx*cscxcancel([(cotx+cscx)/(cotx+cscx)])=cotx*cscx=RHS`