Question

Sec(arctan(-3/5)) is and can you please provide a step by step process thanks!

Answer 1

Let
`x` be some angle so that
`\tan x=-\frac35`.

Recall that

`\sin^2x+\cos^2x=1\implies(\sin^2x)/(\cos^2x)+(\cos^2x)/(\cos^2x)=\frac1{\cos^2x}\implies\tan^2x+1=\sec^2x`

`\implies\sec x=\pm√(1+\tan^2x)=\pm\frac{√(34)}5`

The identity above gives us two possible answers, but they can't both be right:
`\frac{√(34)}5\\eq-\frac{√(34)}5`. The caveat is that
`\arctan` and
`\tan` are not mutual inverse functions. While

`\arctan y=x\implies y=\tan x`

is true, the converse is not. The
`\arctan` function can only return values between
`-\frac\pi2` and
`\frac\pi2`.

To find out which value is correct, we have to use what we know about the
`\tan` function.
`\tan x` is negative only when
`-\frac\pi2<x<0`; over this domain, we expect to have
`\cos x` be positive, which in turn means
`\sec x` should be positive.

So we have

`\sec\left(\arctan\left(-\frac35\right)\right)=\sec x=+\frac{√(34)}5`