#### Answer

$$ y=(2-\sqrt{2}) x+\left[\frac{\sqrt{2} \pi}{4}+\sqrt{2}-1-\frac{\pi}{2}\right]$$

#### Work Step by Step

Given $$y=\csc x-\cot x, \quad x=\frac{\pi}{4}$$
Since $y(\pi/4)= \sqrt{2}-1$ and
\begin{align*}
y'&=-\csc x\cot x+\csc^2 x\\
y'\bigg|_{x=\pi/4}&= (2-\sqrt{2})
\end{align*}
Then the tangent line equation is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y- ( \sqrt{2}-1)}{x-\pi/4}&=(2-\sqrt{2})\\
y-(\sqrt{2}-1)&=(2-\sqrt{2})\left(x-\frac{\pi}{4}\right)
\end{align*}
Hence
$$ y=(2-\sqrt{2}) x+\left[\frac{\sqrt{2} \pi}{4}+\sqrt{2}-1-\frac{\pi}{2}\right]$$