Answer
$$y=\left(x-\frac{\pi}{4}\right)+\ln(1/\sqrt{2})$$
Work Step by Step
Given $$y=\ln (\sin x), \quad x=\frac{\pi}{4}$$
Since at $x=\pi/4$, $f(x) = \ln (1/\sqrt{2}) $ and
$$f'(x)= \frac{\cos x}{\sin x}= \cot x$$
Then $ m= f'(x)\bigg|_{x=\pi/4}=1 $
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x- x_1}&=m\\
\frac{y-\ln (1/\sqrt{2}) }{x- \pi/4}&=1 \\
y&=\left(x-\frac{\pi}{4}\right)+\ln(1/\sqrt{2})
\end{align*}
