જો tan^{-1} frac{a+x}{a} + tan^{-1} frac{a-x} | vedclass

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(C) આપેલ સમીકરણ: $	an^{-1} \left( \frac{a+x}{a} ight) + 	an^{-1} \left( \frac{a-x}{a} ight) = \frac{\pi}{6}$ સૂત્ર $	an^{-1} A + 	an^{-1} B =

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$2\sqrt{3} a^2$

Vedclass · Answer

(C) આપેલ સમીકરણ: $	an^{-1} \left( \frac{a+x}{a} ight) + 	an^{-1} \left( \frac{a-x}{a} ight) = \frac{\pi}{6}$ સૂત્ર $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$ નો ઉપયોગ કરતા: $	an^{-1} \left( \frac{\frac{a+x}{a} + \frac{a-x}{a}}{1 - \left( \frac{a+x}{a} ight) \left( \frac{a-x}{a} ight)} ight) = \frac{\pi}{6}$ $	an^{-1} \left( \frac{\frac{2a}{a}}{1 - \frac{a^2-x^2}{a^2}} ight) = \frac{\pi}{6}$ $	an^{-1} \left( \frac{2}{\frac{a^2 - a^2 + x^2}{a^2}} ight) = \frac{\pi}{6}$ $	an^{-1} \left( \frac{2a^2}{x^2} ight) = \frac{\pi}{6}$ $\frac{2a^2}{x^2} = 	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ $x^2 = 2\sqrt{3} a^2$

જો $\tan^{-1} \frac{a+x}{a} + \tan^{-1} \frac{a-x}{a} = \frac{\pi}{6}$ હોય,તો $x^2 =$

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