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Given, x = a2+b2+a2−b2a2+b2−a2−b2\dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}.

Use componendo and dividendo to prove that :

b2 = 2a2xx2+1\dfrac{2a^2x}{x^2 + 1}

Ratio Proportion

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Answer

Given,

⇒x=a2+b2+a2−b2a2+b2−a2−b2\Rightarrow x = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}

Applying componendo and dividendo:

⇒x+1x−1=a2+b2+a2−b2+a2+b2−a2−b2a2+b2+a2−b2−(a2+b2−a2−b2)⇒x+1x−1=2a2+b22a2−b2⇒x+1x−1=a2+b2a2−b2⇒(x+1)2(x−1)2=a2+b2a2−b2⇒x2+1+2xx2+1−2x=a2+b2a2−b2\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} + \sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} - (\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{a^2 + b^2}}{2\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2}}{\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{a^2 + b^2}{a^2 - b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{a^2 + b^2}{a^2 - b^2}

Applying componendo and dividendo:

⇒x2+1+2x+x2+1−2xx2+1+2x−(x2+1−2x)=a2+b2+a2−b2a2+b2−(a2−b2)⇒2(x2+1)4x=2a22b2⇒x2+12x=a2b2⇒b2=2a2xx2+1.\Rightarrow \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} = \dfrac{a^2 + b^2 + a^2 - b^2}{a^2 + b^2 - (a^2 - b^2)} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{4x} = \dfrac{2a^2}{2b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = \dfrac{a^2}{b^2} \\[1em] \Rightarrow b^2 = \dfrac{2a^2x}{x^2 + 1}.

Hence, proved that b2 = 2a2xx2+1\dfrac{2a^2x}{x^2 + 1}.

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