Question

How do you simplify `1/(1+sin x) + 1/(1-sin x)`?

Answer 1

Let's say your expression is called `E`.

First, multiply the first fraction by `"1-sinx"` and the second by `"1+sinx"`

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

`E = (1 cancel(-sinx) + 1 cancel(+sinx))/((1+sinx) * (1-sinx)) = 2/((1+sinx) * (1-sinx))`

Use the algebraic identity `a^2 - b^2 = (a-b)(a+b)`. In your case,

`a = 1` and
`b = sinx`

As a result, the expression that serves as a denominator will become

`(1+sinx) * (1-sinx) = 1^(2) - (sinx)^(2) = 1 -sin^2x`

Therefore, `E` will be

`E = 2/(1 -sin^2x)`

Remember that `1 - sin^2x = cos^2x`, so the final form of `E` will be

`E = color(green)(2/cos^2x)`