Question

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

• A. $\sqrt{ab}$
• B. $\sqrt{2ab}$
• C. $2ab$
• D. $ab$

Answer 1

The correct option is A $\sqrt{ab}$

Solution :-

$ta{n}^{-1}\left(\frac{a}{x}\right)+ta{n}^{-1}\left(\frac{b}{x}\right)=\frac{\pi }{2}$

$\Rightarrow ta{n}^{-1}\left(\frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{a}{x}\times \frac{b}{x}}\right)=\frac{\pi }{2}$

$\Rightarrow \frac{ax+bx}{x^2-ab}=tan\frac{\pi }{2}$

$\Rightarrow \frac{ax+bx}{x^2-ab}=\infty$

$\Rightarrow \frac{x^2-ab}{ax+bx}=\frac{1}{\infty }$

$\Rightarrow x^2-ab=0$

$\Rightarrow x=\sqrt{ab}$