Q90 / 200 · Manipal (MET) 2014 · Mathematics

If $\tan ^{-1}\left(\frac{a}{x}\right)+\tan ^{-1}\left(\frac{b}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

  • Correct answer
    A
    $\sqrt{a b}$
  • B
    $\sqrt{2 a b}$
  • C
    $2 a b$
  • D
    $a b$

Solution

$\begin{aligned} & \text { Given, } \tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2} \\ & \Rightarrow \quad \tan ^{-1}\left[\frac{\left(\frac{a}{x}+\frac{b}{x}\right)}{1-\frac{a b}{x^2}}\right]=\frac{\pi}{2} \\ & \end{aligned}$ $\begin{array}{rlr}\Rightarrow & \frac{\frac{a+b}{x}}{x^2-a b} & =\tan \frac{\pi}{2} \\ \Rightarrow & x^2 & =a b \\ \Rightarrow & x & =\sqrt{a b}\end{array}$

Hence, the correct answer is option: 'A'.

Relevant Questions