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#