Answer
The equation for the tangent line at $x=-\frac{\pi}{2}$ is $y=-1.$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$$y-f(x_0)=f'(x_0)(x-x_0)$$
Here is $x_0=-\frac{\pi}{2}$ and $f(x)=\sec^3(\frac{\pi}{2}-x)$.
$$f'(x)=(\sec^3(\frac{\pi}{2}-x))'=3\sec^2(\frac{\pi}{2}-x)(\sec(\frac{\pi}{2}-x))'=
3\sec^2(\frac{\pi}{2}-x)\sec(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)(\frac{\pi}{2}-x)'
3\sec^3(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)\cdot(-1)=-3\sec^3(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)$$
Now for $x=-\frac{\pi}{2}$ we have:
$$f(-\frac{\pi}{2})=\sec^2(\frac{\pi}{2}+\frac{\pi}{2})=\sec^3\pi=-1$$
$$f'(-\frac{\pi}{2})=-3\sec^3(\frac{\pi}{2}+\frac{\pi}{2})\tan(\frac{\pi}{2}+\frac{\pi}{2})=-3\sec^3\pi\tan\pi=0$$
So the equation for the tangent line at $x=-\frac{\pi}{2}$ is:
$$y-(-1)=0\cdot(x-(-\frac{\pi}{2}))\Rightarrow y=-1$$
