In this question, why did we turn $|\tan^{-1}\theta|$ to $-\tan^{-1}θ$ in the highlighted step?
Why is there a negative sign? I first thought that the negative sign was because $\frac\pi2 < \theta < \pi$, so $\tan\theta$ is negative on this domain. But if $\tan\theta$ is negative, why do we put a (-) sign and make it positive?
in a similar question [below], we allowed it to be +ve and that seems correct, but i don't understand why in the first question we need to add (-) sign?
In the second line of the equation above, $\sec \theta \lt -1$ is valid in two quadrants: the second and third quadrants.
Substituting $x = \sec \theta$ in the denominator, $\sqrt {\sec^2 \theta - 1} = \sqrt {\tan^2 \theta} = |\tan \theta| = \pm \tan \theta$, but since you're in the range of $\frac {\pi}{2} \lt \theta \lt \pi$, you'll be in the second quadrant; as $\tan \theta$ is negative in the second quadrant, we choose the negative sign, because the absolute value function defines $|x| = -x$, when x is negative.