Hi there!
[tex]\sqrt{\frac{1-cosA}{1+cosA}} =[/tex]
We can begin by multiplying by its conjugate:
[tex]\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}} = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =[/tex]
Simplify using the identity:
[tex]1 - cos^2A = sin^2A[/tex]
[tex]\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =[/tex]
Take the square root of the expression:
[tex]{\frac{sinA}{1+cosA} =[/tex]
Multiply again by the conjugate to get a SINGLE term in the denominator:
[tex]{\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\[/tex]
Simplify:
[tex]{\frac{sinA(1-cosA)}{1-cos^2A} =[/tex]
Use the above trig identity one more:
[tex]{\frac{sinA(1-cosA)}{sin^2A} =[/tex]
Cancel out sinA:
[tex]{\frac{(1-cosA)}{sinA} =[/tex]
Split the fraction into two:
[tex]{\frac{1}{sinA} - \frac{cosA}{sinA} =[/tex]
Recall:
[tex]1/sinA = cscA\\\\cosA/sinA = cotA[/tex]
Simplify:
[tex]\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}[/tex]
