Question

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

Answer 1

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

Explanation:

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

`14cscx`

Explanation:

`(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`