#### Answer

The equation of the tangent line is $y=\sqrt{2}\Big(\dfrac{\pi+4}{4}-x\Big)$

#### Work Step by Step

$y=\csc x$ $;$ $x=\dfrac{\pi}{4}$
Evaluate the derivative of the function given:
$y'=-\csc x\cot x$
Substitute $x=\dfrac{\pi}{4}$ in the derivative found and simplify to obtain the slope of the tangent line at the given point:
$m=-\csc\dfrac{\pi}{4}\cot\dfrac{\pi}{4}=-(\sqrt{2})(1)=-\sqrt{2}$
Substitute $x=\dfrac{\pi}{4}$ in the original expression to obtain the $y$-coordinate of the point given:
$y$-coordinate $=\csc\dfrac{\pi}{4}=\sqrt{2}$
The point is $\Big(\dfrac{\pi}{4},\sqrt{2}\Big)$
The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-\sqrt{2}=-\sqrt{2}\Big(x-\dfrac{\pi}{4}\Big)$
$y-\sqrt{2}=-\sqrt{2}x+\dfrac{\pi\sqrt{2}}{4}$
$y=-\sqrt{2}x+\dfrac{\pi\sqrt{2}}{4}+\sqrt{2}$
$y=-\sqrt{2}x+\dfrac{\pi\sqrt{2}+4\sqrt{2}}{4}$
$y=\sqrt{2}\Big(\dfrac{\pi+4}{4}-x\Big)$
The graph of the function and the tangent line is shown in the answer section.