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Question Number 205990 by MATHEMATICSAM last updated on 04/Apr/24

$$\mathrm{If}x=\frac{\sqrt{3}}{2}\mathrm{then}\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=?$$

Answered by cortano12 last updated on 04/Apr/24

$$\frac{{(\sqrt{1+x}+\sqrt{1-x})}^2}{2x}=\frac{2+2\sqrt{1-x^2}}{2x}$$$$=\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}$$$$=\frac{2}{\sqrt{3}}+\sqrt{\frac{4}{3}-1}=\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\sqrt{3}$$$$$$

Answered by Berbere last updated on 04/Apr/24

$$\frac{\sqrt{3}}{2}=x=sin\left(\frac{\pi }{3}\right)$$$$1+x=2sin\left(\frac{\pi }{3}\right)+1={(sin\left(\frac{\pi }{6}\right)+cos\left(\frac{\pi }{6}\right))}^2$$$$1-x={(cos\left(\frac{\pi }{6}\right)-sin\left(\frac{\pi }{6}\right))}^2$$$$\iff \frac{2cos\left(\frac{\pi }{6}\right)}{2sin\left(\frac{\pi }{6}\right)}=\frac{1}{\tan \left(\frac{\pi }{6}\right)}=\sqrt{3}$$

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