Question

How do we show that `(sinx - cosx)/sinx + (cosx -sinx)/cosx = 2 - secxcscx`?

Answer 1

Put the left-hand side on a common denominator and convert the right hand side to sine and cosine (using `secx = 1/cosx` and `cscx = 1/sinx)`.

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

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

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

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

Now put the right-hand side on a common denominator and use the identity `sin^2x + cos^2x = 1` to simplify the left-hand side.

`(2cosxsinx - 1)/(sinxcosx) = (2(sinxcosx) - 1)/(sinxcosx)`

`(2cosxsinx - 1)/(sinxcosx) = (2sinxcosx - 1)/(sinxcosx)`

This is obviously true, so we have proven the identity.

Hopefully this helps!