Question

How do you simplify `\frac { 1} { \sin x + 1} + \frac { 1} { \sin x - 1}`?

Answer 1

`(2sinx)/(sin^2x-1)`

Explanation:

First get a common denominator by multiplying the first term by `1` in the form of `(sinx-1)/(sinx-1)` and the second in the form of `(sinx+1)/(sinx+1)`

`(1(sinx-1))/((sinx+1)(sinx-1))+(1(sinx+1))/((sinx-1)(sinx+1))`

Multiply out

`(sinx-1)/(sin^2x-1)+(sinx+1)/(sin^2x-1)`

And combine

`(sinx-1+sinx+1)/(sin^2x-1)`

Combine like terms

`(2sinx)/(sin^2x-1)`

And you're done!

Answer 2

• 2 tan x.sec x

Explanation:

Put the 2 fractions having common denominator
`f(x) = (2sin x)/(sin^2 x - 1) = (2sin x)/-(1 - sin^2 x) = - (2sin x)/cos^2 x =`
`f(x) = - 2tan x(1/cos x) = - 2tan x.sec x`