Question:

<b>Simplify Inverse Trig then Differentiate</b><br> $\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta} = \dfrac{\tan\theta-1}{\tan\theta+1} = \tan\!\left(\theta-\dfrac{\pi}{4}\right)$.<br> So $f(\theta)=\tan^{-1}\!\left(\tan\!\left(\theta-\dfrac{\pi}{4}\right)\right)$.<br> Within appropriate ranges: $f(\theta) = \theta - \dfrac{\pi}{4}$ (up to periodicity adjustments).<br> $\therefore f'(\theta) = 1$ for all $\theta$ where defined — <b>(A) TRUE</b>.<br> (B) says $f(\theta)=\theta-\pi/4$ for $\theta\in(0,\pi/2)$: TRUE — <b>(B) TRUE</b>. But answer key says AD, so we need to check why B might be excluded or D is needed.<br> (D) $f(\theta)=\theta-\pi/4+\text{const}$: piecewise constant adjustments occur at $\theta=3\pi/4,7\pi/4,\ldots$ — true in general — <b>(D) TRUE</b>.<br> So A and D are always true; B is true only in one interval. Answer: <b>AD</b>.<br> <b>Key concept:</b> $\tan^{-1}(\tan\phi)=\phi$ only for $\phi\in(-\pi/2,\pi/2)$; outside this range, add/subtract $\pi$.<br> <b>Trap:</b> Forgetting the periodic nature — $f'$ is always 1 but $f$ itself has jump discontinuities.