MCQ
If $\cot ^{-1} \alpha+\cot ^{-1} \beta=\cot ^{-1} x$, then $x=$
  • A
    $\alpha+\beta$
  • B
    $\alpha-\beta$
  • C
    $\frac{1+\alpha \beta}{\alpha+\beta}$
  • $\frac{\alpha \beta-1}{\alpha+\beta}$

Answer

Correct option: D.
$\frac{\alpha \beta-1}{\alpha+\beta}$
(D) $\cot ^{-1} \alpha+\cot ^{-1} \beta=\cot ^{-1} x$
$\Rightarrow \cot ^{-1}\left(\frac{\alpha \beta-1}{\alpha+\beta}\right)=\cot ^{-1} x$ $\ldots\left[\because \cot ^{-1} x+\cot ^{-1} y=\cot ^{-1}\left(\frac{x y-1}{x+y}\right)\right]$
$\Rightarrow x=\frac{\alpha \beta-1}{\alpha+\beta}$

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