Question

Sinx/1+cosx+cosx/sinx=1/sinx?

Answer 1

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)sin^2x+cos^2x=1`

`"consider the left side"`

`sinx/(1+cosx)+cosx/sinx`

`"express as a single fraction"`

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

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

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

`=1/sinx=" right side"`

Answer 2

`1/sinx`

Explanation:

We have the following

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

After getting common denominators, we would get

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

`=>(sin^2x+cosx(1+cosx))/((sinx)(1+cosx))=(1+cosx)/(sinx(1+cosx))`

Further simplifying, we get

`(color(blue)(sin^2x+cos^2x)+cosx)/((1+cosx)(sinx))=(1+cosx)/(sinx(1+cosx))`

What I have in blue is the Pythagorean Identity, which just simplifies to `1`. Now, we have

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

We have some terms to cancel:

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

and we're just left with

`1/sinx=1/sinx`

We have essentially proven this "identity".

Hope this helps!