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).