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

3 Answers
Mar 15, 2018

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)

Mar 15, 2018

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!

Mar 15, 2018

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