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प्रश्न
Given `x = (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: `b^2 = (2a^2x)/(x^2 + 1)`
उत्तर
Given: `x/(1) = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`
Applying componendo and dividendo both sides, we have
`(x + 1)/(x - 1) = (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)`
`\implies (x + 1)/(x - 1) = (2sqrt(a^2 + b^2))/(2sqrt(a^2 + b^2)`
`\implies (x + 1)/(x - 1) = (sqrt(a^2 + b^2))/(sqrt(a^2 + b^2)`
Squaring both sides, we have
`\implies (x + 1)^2/(x - 1)^2 = (a^2 + b^2)/(a^2 - b^2)`
`\implies (x^2 + 1 + 2x)/(x^2 + 1 - 2x) = (a^2 + b^2)/(a^2 - b^2)`
Applying componendo and dividendo both sides, we get
`\implies (x^2 + 1 + 2x + x^2 + 1 - 2x)/(x^2 + 1 + 2x - x^2 - 1 + 2x) = (a^2 + b^2 + a^2 - b^2)/(a^2 + b^2 - a^2 + b^2)`
`\implies (2x^2 + 2)/(4x) = (2a^2)/(2b^2)`
`\implies (x^2 + 1)/(2x) = a^2/b^2`
`\implies b^2 = (2a^2x)/(x^2 + 1)`
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