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  • If   <img alt="\mathrm{tan^{-1}\left ( \frac{2a}{x} \right )+tan^{-1}\left ( \frac{2b}{x} \right )=\frac{\pi}{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Btan%5E%7B-1%

If   \mathrm{tan^{-1}\left ( \frac{2a}{x} \right )+tan^{-1}\left ( \frac{2b}{x} \right )=\frac{\pi}{2}}, then the value of \mathrm{ x} is

Option: 1

\mathrm{ ab }


Option: 2

\mathrm{\pm 2\sqrt{ab} }


Option: 3

\mathrm{ \pm \sqrt{ab} }


Option: 4

\mathrm{ \frac{a}{b} }


Answers (1)

best_answer

Option (b) \mathrm{\pm 2\sqrt{ab} }

Given that,

\mathrm{ tan^{-1}\left ( \frac{2a}{x} \right )+tan^{-1}\left ( \frac{2b}{x} \right )=\frac{\pi}{2} }

\mathrm{ tan^{-1}\left ( \frac{2a}{x} \right )=\frac{\pi}{2}-tan^{-1}\left ( \frac{2b}{x} \right ) }

\mathrm{ tan^{-1}\left ( \frac{2a}{x} \right )=cot^{-1}\left ( \frac{2b}{x} \right ) }

\mathrm{tan^{-1}\left ( \frac{2a}{x} \right )=tan^{-1}\left ( \frac{x}{2b} \right ) }

\mathrm{\frac{2a}{x}=\frac{x}{2b} }

\mathrm{ x^{2}=4ab }

\mathrm{ x=\pm 2\sqrt{ab} }

Posted by

Suraj Bhandari

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