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Question: If \(\cot ^ { - 1 } \alpha + \cot ^ { - 1 } \beta = \cot ^ { - 1 } x\) then \(x =\)...

If cot⁡−1α+cot⁡−1β=cot⁡−1x\cot ^ { - 1 } \alpha + \cot ^ { - 1 } \beta = \cot ^ { - 1 } x then x=x =

A

α+β\alpha + \beta

B

α−β\alpha - \beta

C

1+αβα+β\frac { 1 + \alpha \beta } { \alpha + \beta }

D

αβ−1α+β\frac { \alpha \beta - 1 } { \alpha + \beta }

Answer

αβ−1α+β\frac { \alpha \beta - 1 } { \alpha + \beta }

Explanation

Solution

Given that cot⁡−1α+cot⁡−1β=cot⁡−1x\cot ^ { - 1 } \alpha + \cot ^ { - 1 } \beta = \cot ^ { - 1 } x

⇒cot⁡−1(αβ−1α+β)=cot⁡−1x⇒x=αβ−1α+β\Rightarrow \cot ^ { - 1 } \left( \frac { \alpha \beta - 1 } { \alpha + \beta } \right) = \cot ^ { - 1 } x \Rightarrow x = \frac { \alpha \beta - 1 } { \alpha + \beta } .