Q.Show that $2\tan^{-1}(-3)=\frac{-\pi}{2}+\tan^{-1}\left(\frac{-4}{3}\right)$.

Let $\theta=\tan^{-1}(-3)$, so $\tan\theta=-3$ and $\theta\in\left(-\frac{\pi}{2},0\right)$; hence $2\theta\in(-\pi,0)$.

Double angle:

Fix the branch: numerically $\theta\approx-1.249$, so $2\theta\approx-2.498\in\left(-\pi,-\frac{\pi}{2}\right)$, while $\tan^{-1}\frac{3}{4}\approx0.6435\in\left(0,\frac{\pi}{2}\right)$. These have the same tangent and differ by one period, so

Rewrite: for $a>0$, $\tan^{-1}a+\tan^{-1}\frac{1}{a}=\frac{\pi}{2}$, so $\tan^{-1}\frac{3}{4}=\frac{\pi}{2}-\tan^{-1}\frac{4}{3}$. Therefore …

The idea

We cannot simply take $\tan^{-1}$ of $\tan 2\theta$, because $2\theta$ may fall outside the principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. So we compute $\tan 2\theta$, locate $2\theta$ exactly, and correct by a multiple of $\pi$.

Step 1 — Set up

Let $\theta=\tan^{-1}(-3)$. Then $\tan\theta=-3$, and since the argument is negative, $\theta\in\left(-\frac{\pi}{2},0\right)$. Doubling, $2\theta\in(-\pi,0)$.

Step 2 — Tangent of the double angle

Step 3 — Place $2\theta$ correctly

Numerically $\theta\approx-1.249$, so $2\theta\approx-2.498$, which lies in $\left(-\pi,-\frac{\pi}{2}\right)$. The principal value $\tan^{-1}\frac{3}{4}\approx0.6435$ lies in $\left(0,\frac{\pi}{2}\right)$. These two angles share the same tangent and differ by exactly one period $\pi$, so

Step 4 — Use the complementary identity …

Method: Rewriting $2\tan^{-1}a$ with a branch correction

When you double an inverse tangent whose value is large or negative, the doubled angle can leave the principal range, so the clean identity needs a $\pm\pi$ correction. This is the general technique for such "show that" identities.

Steps

Step 1: Name the angle and locate $2\theta$.

Let $\theta=\tan^{-1}a$. If $a<0$ then $\theta\in\left(-\tfrac{\pi}{2},0\right)$, so $2\theta\in(-\pi,0)$ — already a warning that $2\theta$ may fall below $-\tfrac{\pi}{2}$.

Step 2: Compute the tangent of the double angle.

Step 3: Correct the branch. …

Common Mistakes

Mistake 1: Applying $2\tan^{-1}x=\tan^{-1}\tfrac{2x}{1-x^2}$ with no branch correction.

Why it's wrong: the clean identity needs $|x|<1$; for $x=-3$ the doubled angle leaves $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, so a $\pm\pi$ term is required. Correct approach: locate $2\theta$ (here $\approx-2.498\in(-\pi,-\tfrac{\pi}{2})$) and write $2\theta=\tan^{-1}\tfrac34-\pi$.

Mistake 2: Forgetting the domain of $2\theta$ and picking the wrong sign of $\pi$. …