How do you simplify #(5+9cosx) / (sinx) + (9sinx) / (1+cosx)#?

Question

2370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-17T06:01:44

#(5+9cosx)/(sinx) + (9sinx)/(1+cosx) = 14cscx#

We can start off this problem by getting a common denominator.

#(5+9cosx)/(sinx) + (9sinx)/(1+cosx)#

#= ((5+9cosx)(1+cosx) + 9sin^2 x)/(sinx(1+cosx))#

Multiplying everything out gives us

#=(5+5cos x + 9cosx + 9cos^2 x + 9 sin^2 x)/(sinx(1+cosx))#

Factoring out a #9# yields

#=(5+14cosx + 9(cos^2 x + sin^2 x))/(sinx(1+cosx))#

#=(5+14cosx +9 )/(sinx(1+cosx))#

Adding like terms of #9# and #5#, which equals to #14# results in

#=(14+14cosx)/(sinx(1+cosx))#

Factoring out a #14# yields

#=(14cancel((1+cosx)))/(sinxcancel((1+cosx))) =14/(sinx) = 14cscx#

Answer 2 · 2016-07-17T06:16:08

#14cscx#

#(5+9cosx)/sinx+(9sinx)/((1+cosx)#

#=(5+9cosx)/sinx+(9sinx(1-cosx))/((1+cosx)(1-cosx))#

#=(5+9cosx)/sinx+(9sinx(1-cosx))/(1-cos^2x)#

#=(5+9cosx)/sinx+(9sinx(1-cosx))/sin^2x#

#=(5+9cosx)/sinx+(9(1-cosx))/sinx#

#=(5/sinx+(9cosx)/sinx)+(9/sinx-(9cosx)/sinx)#

#=5cscx+9cotx+9cscx-9cotx#

#=14cscx#