Please help me with a trig proof

[wanocuni]

How do you solve this proof?
This is how far I got, but I don't know what to do past this unless I did something wrong.

 

[BigBeachBanana]

The step where you inverted the fractions is incorrect.

Combine [imath]\frac{\cos^2x}{\sin^2x}-1[/imath] into one fraction first, then invert the fraction.

 

[pka]

[imath]\dfrac{1+\cot^2(x)}{\cot^2(x)-1}=\dfrac{1+\frac{\cos^2(x)}{\sin^2(x)}}{\frac{\cos^2(x)}{\sin^2(x)}-1}=\dfrac{\sin^2(x)+\cos^2(x)}{\cos^2(x)-\sin^2(x)}[/imath]

 

[wanocuni]

The step where you inverted the fractions is incorrect.
View attachment 30869
Combine [imath]\frac{\cos^2x}{\sin^2x}-1[/imath] into one fraction first, then invert the fraction.

OMG thanks so much!
I got it now

 

[Steven G]

The reciprocal of (a + b) written as rec(a +b) does NOT equal rec(a) + rec(b)

\(\displaystyle rec(a+b) = \dfrac{1}{a+b} \neq rec(a) + rec(b) = \dfrac{1}{a} + \dfrac{1}{b} = \dfrac {a+b}{ab}\)

This is what BBB is saying.