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
    $\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1-3\text{a}^2}\Big)$
  • $\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$

Answer

Correct option: D.
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$

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