First slide

Properties of ITF

Question

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

Moderate

A

$\sqrt{ab}$
B

$\sqrt{2 ab}$
C

2ab
D

ab

Solution

$\begin{array}{l} \tan^{- 1} (\frac{a}{x}) + \tan^{- 1} (\frac{b}{x}) = \frac{π}{2} \\ \Rightarrow \tan^{- 1} (\frac{\frac{a}{x} + \frac{b}{x}}{1 - \frac{ab}{x^{2}}}) = \frac{π}{2} \\ \Rightarrow \frac{\frac{a}{x} + \frac{b}{x}}{1 - \frac{ab}{x^{2}}} = \tan \frac{π}{2} \\ \Rightarrow 1 - \frac{ab}{x^{2}} = 0 \\ \Rightarrow x^{2} = ab \\ \Rightarrow x = \sqrt{ab} \end{array}$

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