Question

How can I solve question b). ?

Answer 1

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:

$\text{b)} \frac{\sin(a)}{\sin(a)-\cos(a)} - \frac{\cos(a)}{\cos(a)-\sin(a)} = \frac{1+\cot^2 (a)}{1-\cot^2 (a)}$

You want to verify this identity.

$\frac{\sin(a)(\cos(a)-\sin(a))}{(\sin(a)-\cos(a))(\cos(a)-\sin(a))} - \frac{\cos(a)(\sin(a)-\cos(a))}{(\sin(a)-\cos(a))(\cos(a)-\sin(a))} = \frac{1+\cot^2 (a)}{1-\cot^2 (a)}$

The common denominator is

$(\sin(a)-\cos(a))(\cos(a)-\sin(a))= \boxed{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}$

Solving the first and second numerator:

$\sin(a)(\cos(a)-\sin(a))=\sin(a)\cos(a)-\sin^2(a)$

$\cos(a)(\sin(a)-\cos(a))= \cos(a)\sin(a)-\cos^2(a)$

Now we have

$\frac{ \sin(a)\cos(a)-\sin^2(a) -(\cos(a)\sin(a)-\cos^2(a))}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}$

$\frac{ -\sin^2(a) +\cos^2(a)}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}$

Once

$-\sin^2(a) +\cos^2(a) = \cos(2a)$

$2\cos (a)\sin(a) = \sin(2a)$

Also, consider the identity:

$\boxed{\sin^2(a)+\cos^2(a)=1}$

$\frac{ -\sin^2(a) +\cos^2(a)}{2\cos (a)\sin(a)-\cos ^2(a)-\sin ^2(a)}=\boxed{\frac{ \cos(2a)}{\sin(2a)-1}}$

That last claim is true.