I need to prove that the following is true. Thanks

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

I recall this question causing all kinds of problems when I was still teaching.

it requires a little "trick"

L.S.
=cosx/(1-sinx) multiply top and bottom by 1+sinx, (creating the difference of square pattern)
=cosx(1+sinx)/[(1-sinx)(1+sinx)]
=cosx(1+sinx)/(1-sin^2x)
=cosx(1+sinx)/cos^2x divide by cosx
= (1+sinx)/cosx
= R.S.

Wow !!

To prove that (cosx / 1-sinx) is equal to (1+sinx / cosx), you can use a trigonometric identity and some algebraic manipulation.

Let's start with the left-hand side (L.H.S.):

L.H.S. = cosx / (1-sinx)

Now, multiply both the numerator and denominator by (1+sinx):

L.H.S. = (cosx * (1+sinx)) / ((1-sinx) * (1+sinx))

In the denominator, you can use the difference of squares pattern: (a^2 - b^2) = (a+b)(a-b). In this case, a is 1 and b is sinx. So:

L.H.S. = (cosx * (1+sinx)) / ((1^2 - sin^2x))

Simplifying further:

L.H.S. = (cosx * (1+sinx)) / (1 - sin^2x)

Since sin^2x + cos^2x = 1 (based on the Pythagorean identity), we can substitute cos^2x for 1 - sin^2x:

L.H.S. = (cosx * (1+sinx)) / cos^2x

Dividing the numerator and denominator by cosx:

L.H.S. = (1+sinx) / cosx

This is equivalent to the right-hand side (R.H.S.), which is our desired expression. Therefore, the given equation is true.

So, by using the trigonometric identity sin^2x + cos^2x = 1 and applying algebraic manipulation, we have proven that (cosx / 1-sinx) is equal to (1+sinx / cosx).