Question

Can someone help on this trig proof?
(1+cosx)/(1-cosx)=(cscx+cotx)^2
I started by eliminating cos from the denominator
Multiply by ((1+cosx)/(1+cosx))
(1+2cosx+cos^2x)/(1-cos^2x) use identity to change 1-cos^2x to sin^2x so now I have
(1+2cosx+cos^2x)/sin^2x
I am stuck. Did I do a step wrong?

Answer 1

Are you allowed to play with both sides at the same time, or not?

Answer 2

no I have to work from one side to equal the other

Answer 3

if you can reverse engineer it I think that is vaild :)

Answer 4

\(\large\color{black}{ \displaystyle \frac{ 1+\cos x }{1-\cos x} =\left(\csc x + \cot x\right)^2 }\)

\(\large\color{black}{ \displaystyle \frac{ 1+\cos x }{1-\cos x} =\left(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right)^2 }\)

\(\large\color{black}{ \displaystyle \frac{ 1+\cos x }{1-\cos x} =\left(\frac{1+\cos x}{\sin x}\right)^2 }\)

\(\large\color{black}{ \displaystyle \frac{ 1+\cos x }{1-\cos x} =\frac{(1+\cos x)^2}{\sin^2x} }\)

\(\large\color{black}{ \displaystyle \frac{ 1+\cos x }{1-\cos x} =\frac{(1+\cos x)^2}{1-\cos^2x} }\)

Answer 5

then, the last step (on the right side), is to divide by \(1+\cos x\) on top and bottom.

Answer 6

Oh that is so much more simplistic! Beautiful Solomon!

Answer 7

Anytime:)