Q.Show that $\tan^{-1}\left(\dfrac{1}{2}\right) + \tan^{-1}\left(\dfrac{2}{11}\right) = \tan^{-1}\left(\dfrac{3}{4}\right)$.

Using $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab}$ with $a=\tfrac12,\ b=\tfrac{2}{11}$ (so $ab=\tfrac{1}{11}<1$) gives $\tan^{-1}\tfrac34$.

Step 1 — Addition formula. For $ab < 1$,

$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\frac{a+b}{1-ab}\right).$

Here $a = \dfrac12,\ b = \dfrac{2}{11}$, and $ab = \dfrac{2}{22} = \dfrac{1}{11} < 1$, so the formula applies.

Step 2 — Numerator $a+b$.

$\frac12 + \frac{2}{11} = \frac{11 + 4}{22} = \frac{15}{22}.$

Step 3 — Denominator $1-ab$.

$1 - \frac{1}{11} = \frac{10}{11}.$

Step 4 — Combine. …