Why does $-\sec(x)$ become positive $1/\cos(x)$? I thought that the negative would carry and it would be $-1/\cos(x)$.
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$\begingroup$You are right. Just by looking at the definition of $\sec(x)$, one can clearly see that $$\sec (x) = \frac{1}{\cos (x)}$$
But don't just take my word for it. Your equation itself will tell you if it's right or wrong.
$$ -\sec (x) = \frac{1}{\cos (x)} \implies -\frac{1}{\cos (x)} = \frac{1}{\cos (x)} \implies \large{-1 =1} \implies {\text{False}}$$
Now, there is a result similar to this which is true, $$ \sec (-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos(x)}$$
Why did that happen? Because $ \cos(-x) = \cos(x)$.
Again don't just take my word for it. I can prove it using the identity $\cos(A \pm B) = \cos(A)\times\cos(B) \mp \sin(A)\times\sin(B)$. $$\cos(-x) = \cos (0-x) = \cos 0\times\cos x + \sin 0\times\sin x = 1\times\cos x + 0\times\sin x = \cos x$$
Never confuse $\sec(-x)$ with $-\sec(x)$. Always remember that $$ \sec(-x) = +\sec(x) \quad\forall x\in\mathbb{R} $$
$\ddot\smile$ Hope this helps
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