Answer
$$y = \frac{1}{3}x + \frac{1}{2}\ln \left( {\frac{3}{2}} \right) - \frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \sqrt {1 + {{\sin }^2}x} ,{\text{ }}\left( {\frac{\pi }{4},\ln \sqrt {\frac{3}{2}} } \right) \cr
& {\text{Differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \sqrt {1 + {{\sin }^2}x} } \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 + {{\sin }^2}x} \right)} \right] \cr
& f'\left( x \right) = \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 + {{\sin }^2}x} \right)} \right] \cr
& f'\left( x \right) = \frac{1}{2}\left( {\frac{{0 + 2\sin x\cos x}}{{1 + {{\sin }^2}x}}} \right) \cr
& f'\left( x \right) = \frac{{\sin x\cos x}}{{1 + {{\sin }^2}x}} \cr
& {\text{Calculate the slope at the given point }}\left( {\frac{\pi }{4},\ln \sqrt {\frac{3}{2}} } \right) \cr
& m = f'\left( {\frac{\pi }{4}} \right) = \frac{{\sin \left( {\pi /4} \right)\cos \left( {\pi /4} \right)}}{{1 + {{\sin }^2}\left( {\pi /4} \right)}} \cr
& m = \frac{{1/2}}{{1 + 1/2}} = \frac{1}{3} \cr
& {\text{Find the equation of the tangent line at }}\left( {\frac{\pi }{4},\ln \sqrt {\frac{3}{2}} } \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \ln \sqrt {\frac{3}{2}} = \frac{1}{3}\left( {x - \frac{\pi }{4}} \right) \cr
& y - \ln \sqrt {\frac{3}{2}} = \frac{1}{3}x - \frac{\pi }{{12}} \cr
& y = \frac{1}{3}x + \ln \sqrt {\frac{3}{2}} - \frac{\pi }{{12}} \cr
& y = \frac{1}{3}x + \frac{1}{2}\ln \left( {\frac{3}{2}} \right) - \frac{\pi }{{12}} \cr
& \cr
& {\text{Graph}} \cr} $$
