Inverse Trig Equation Comparison: Arcsec vs Arctan

[IniquiTrance]

Homework Statement

I have knocked a problem down to:

$$-arcsec \frac{3x}{4}$$

but my textbook gives the answer as:

$$arctan \frac{4}{\sqrt{9x^2 - 16}}$$

Are these somehow equivalent?

Thanks

Homework Equations

The Attempt at a Solution

 

[Mark44]

I don't think so. However, I get arccsc (3x/4) being equal to arctan(4/sqrt(9x^2 - 16))