Answer:
It was not my intention to post that answer, as it does not solve the question, but hope it helps somehow.
Step-by-step explanation:
[tex]$\text{b)} \frac{\sin(a)}{\sin(a)-\cos(a)} - \frac{\cos(a)}{\cos(a)-\sin(a)} = \frac{1+\cot^2 (a)}{1-\cot^2 (a)} $[/tex]
You want to verify this identity.
[tex]$\frac{\sin(a)(\cos(a)-\sin(a))}{(\sin(a)-\cos(a))(\cos(a)-\sin(a))} - \frac{\cos(a)(\sin(a)-\cos(a))}{(\sin(a)-\cos(a))(\cos(a)-\sin(a))} = \frac{1+\cot^2 (a)}{1-\cot^2 (a)} $[/tex]
The common denominator is
[tex](\sin(a)-\cos(a))(\cos(a)-\sin(a))= \boxed{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}[/tex]
Solving the first and second numerator:
[tex]\sin(a)(\cos(a)-\sin(a))=\sin(a)\cos(a)-\sin^2(a)[/tex]
[tex]\cos(a)(\sin(a)-\cos(a))= \cos(a)\sin(a)-\cos^2(a)[/tex]
Now we have
[tex]$\frac{ \sin(a)\cos(a)-\sin^2(a) -(\cos(a)\sin(a)-\cos^2(a))}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}$[/tex]
[tex]$\frac{ -\sin^2(a) +\cos^2(a)}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}$[/tex]
Once
[tex]-\sin^2(a) +\cos^2(a) = \cos(2a)[/tex]
[tex]2\cos (a)\sin(a) = \sin(2a)[/tex]
Also, consider the identity:
[tex]\boxed{\sin^2(a)+\cos^2(a)=1}[/tex]
[tex]$\frac{ -\sin^2(a) +\cos^2(a)}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}=\boxed{\frac{ \cos(2a)}{\sin(2a)-1}}$[/tex]
That last claim is true.
