MCQ
$3\tan^{-1} a$ is equal to:
- A$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1+3\text{a}^2}\Big)$
- B$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1+3\text{a}^2}\Big)$
- C$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1-3\text{a}^2}\Big)$
- ✓$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$