If $z=\sqrt{3}+i=(a+ib)(c+id)$, then find $\tan^{-1}\dfrac{b}{a}+\tan^{-1}\dfrac{d}{c}$

Question

If $z=\sqrt{3}+i=(a+ib)(c+id)$, then $\tan^{-1}\dfrac{b}{a}+\tan^{-1}\dfrac{d}{c}=$ _______________

$$
\arg z=\theta,\quad\arg(a+ib)=\theta_1,\quad\arg(c+id)=\theta_2\implies\arg(z)=\theta=\theta_1+\theta_2=\frac{\pi}{6}\\
\tan\theta=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}},\quad\tan\theta_1=\frac{b}{a},\quad\tan\theta_2=\frac{d}{c}\\
\implies\theta_1=n\pi+\tan^{-1}\frac{b}{a},\quad\theta_2=m\pi+\tan^{-1}\frac{d}{c}\\
\theta=\frac{\pi}{6}=(n+m)\pi+\tan^{-1}\frac{b}{a}+\tan^{-1}\frac{d}{c}\\
\tan^{-1}\frac{b}{a}+\tan^{-1}\frac{d}{c}=\frac{\pi}{6}-(n+m)\pi=\frac{\pi}{6}+k\pi
$$
Is it the right way to prove that $\tan^{-1}\dfrac{b}{a}+\tan^{-1}\dfrac{d}{c}=n\pi+\dfrac{\pi}{6}$ ?

Answer 1

I think that a simpler way would be\begin{align}\tan\left(\arctan\left(\frac ba\right)+\arctan\left(\frac dc\right)\right)&=\frac{\frac ba+\frac dc}{1-\frac ba\times\frac dc}\\&=\frac{bc+ad}{ac-bd}\\&=\frac1{\sqrt3}.\end{align}