Question

Question #c175d

Answer 1

The answer is `=(1-cosx)/sinx`

Explanation:

We need

`sin^2x+cos^2x=1`

Therefore,

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

`=((1+sinx)^2-2cosx(1+sinx)+cos^2x)/((1+sinx)^2-cos^2x)`

`=(1+2sinx+sin^2x-2cosx-2cosxsinx+cos^2x)/(1+2sinx+sin^2x-cos^2x)`

`=(2+2sinx-2cosx-2cosxsinx)/(2sin^2x+2sinx)`

`=(2(1+sinx)-2cosx(1+sinx))/(2sinx(1+sinx))`

`=(cancel((1+sinx))(1-cosx))/(sinxcancel((1+sinx)))`

`=(1-cosx)/sinx`

Answer 2

tan (x/2)

Explanation:

`N/D = (1 + sin x - cos x)/(1 + sin x + cos x)`
Develop the numerator N, and denominator D -->
`N = (1 - cos x) + sin x = 2sin^2 (x/2) + sin x =`
`N = 2sin^2 (x/2) + 2sin (x/2).cos (x/2) =`
`N = 2sin (x/2)(sin (x/2) + cos (x/2))`
Same development:
`D = (1 + cos x) + sin x = 2cos^2 (x/2) + 2sin (x/2).cos (x/2)`
`D = 2cos (x/2)(cos (x/2) + sin (x/2))`
Finally:
`N/D = (2sin (x/2))/(2cos (x/2)) = tan (x/2)`