Answer
See proof
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln \sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \csc x \cr
& {\text{Rewrite the left side using radical properties}} \cr
& \frac{d}{{dx}}\left[ {\ln {{\left( {\frac{{1 - \cos x}}{{1 + \cos x}}} \right)}^{1/2}}} \right] \cr
& {\text{Use the logarithmic property }}\ln {u^n} = n\ln u \cr
& \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {\frac{{1 - \cos x}}{{1 + \cos x}}} \right)} \right] \cr
& {\text{Differentiating the left side}}{\text{, use }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {\frac{{1 - \cos x}}{{1 + \cos x}}} \right)} \right] = \frac{1}{2}\left( {\frac{{1 + \cos x}}{{1 - \cos x}}} \right)\frac{d}{{dx}}\left[ {\frac{{1 - \cos x}}{{1 + \cos x}}} \right] \cr
& {\text{Using the quotient rule}} \cr
& = \frac{1}{2}\left( {\frac{{1 + \cos x}}{{1 - \cos x}}} \right)\left( {\frac{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)' - \left( {1 - \cos x} \right)\left( {1 + \cos x} \right)'}}{{{{\left( {1 + \cos x} \right)}^2}}}} \right) \cr
& {\text{Computing the derivatives}} \cr
& = \frac{1}{2}\left( {\frac{1}{{1 - \cos x}}} \right)\left( {\frac{{\left( {1 + \cos x} \right)\left( {\sin x} \right) - \left( {1 - \cos x} \right)\left( { - \sin x} \right)}}{{1 + \cos x}}} \right) \cr
& {\text{Simplifying}} \cr
& = \frac{1}{2}\left( {\frac{1}{{1 - \cos x}}} \right)\left( {\frac{{\sin x + \sin x\cos x + \sin x - \sin x\cos x}}{{1 + \cos x}}} \right) \cr
& = \frac{1}{2}\left( {\frac{1}{{1 - \cos x}}} \right)\left( {\frac{{2\sin x}}{{1 + \cos x}}} \right) \cr
& = \frac{{\sin x}}{{1 - {{\cos }^2}x}} \cr
& {\text{Use the identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \frac{{\sin x}}{{{{\sin }^2}x}} \cr
& = \frac{1}{{\sin x}} \cr
& = \csc x \cr
& {\text{Therefore}}{\text{, the statement has been verified.}} \cr} $$
