KnowledgeBoat Logo
|

Mathematics

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

53 Likes

Answer

Given,

x=a2+b2+a2−b2a2+b2−a2−b2x = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}

By 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\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}}

On squaring both sides,

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

Again applying componendo and dividendo,

⇒x2+1+2x+x2+1−2xx2+1+2x−x2−1+2x=a2+b2+a2−b2a2+b2−a2+b2⇒2x2+24x=2a22b2⇒x2+12x=a2b2⇒b2=2a2xx2+1.\Rightarrow \dfrac{x^2 + 1 + \cancel{2x} + x^2 + 1 - \cancel{2x}}{\cancel{x^2} + \cancel{1} + 2x - \cancel{x^2} - \cancel{1} + 2x} = \dfrac{a^2 + \cancel{b^2} + a^2 - \cancel{b^2}}{\cancel{a^2} + b^2 - \cancel{a^2} + b^2} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{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.b^2 = \dfrac{2a^2x}{x^2 + 1}.

Answered By

25 Likes

Related Questions