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प्रश्न
If y = `(sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`, show that 3by2 - 2ay + 3b = 0.
उत्तर
We have
`y/(1) = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`
Applying componendo and dividendo
`(y + 1)/(y - 1) = (sqrt(a + 3b) + sqrt(a - 3b) + sqrt(a + 3b) - sqrt(a - 3b))/(sqrt(a + 3b) + sqrt(a - 3b) - sqrt(a + 3b) + sqrt(a - 3b))`
`(y + 1)/(y - 1) = (2sqrt(a + 3b))/(2sqrt(a - 3b)`
Squaring both side
`((y + 1)^2)/((y - 1)^2) = (a + 3b)/(a - 3b)`
⇒ `(y^2 + 1 + 2y)/(y^2 + 1 - 2y) = (a + 3b)/(a - 3b)`
Again applying componendo and dividendo
⇒ `(y^2 + 1 + 2y + y^2 + 1 - 2y)/(y^2 + 1 + 2y - y62 - 1 + 2y) = (a + 3b + a - 3b)/(a + 3b - a + 3b)`
⇒ `(2(y^2 + 1))/(4y) = (2a)/(6b)`
⇒ 3by2 + 3b = 2ay
⇒ 3by2 - 2ay + 3 = 0.
Hence proved.
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