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प्रश्न
If `sqrt(1 - x^2) + sqrt(1 - y^2) = "a"(x - y)`, prove that `"dy"/"dx" = sqrt((1 - y^2)/(1 - x^2)`
उत्तर
Given that: `sqrt(1 - x^2) + sqrt(1 - y^2) = "a"(x - y)`
Put x = sin θ and y = sin Φ.
∴ θ = sin–1x and Φ = sin–1y
`sqrt(1 - sin^2theta) + sqrt(1 - sin^2phi)` = a(sin θ – sin Φ)
⇒ `sqrt(cos^2theta) + sqrt(cos^2phi)` = a(sin θ – sin Φ)
⇒ cos θ + cos Φ = a(sin θ – sin Φ)
⇒ `(cos theta + cos phi)/(sin theta - sin phi)` = a
⇒ `(2 cos (theta + phi)/2 * cos (theta - phi)/2)/(2cos (theta + phi)/2 * sin (theta - phi)/2)` = a ......`[(because cos "A" + cos "B" = 2cos ("A" + "B")/2 * cos ("A" - "B")/2),(sin"A" - sin"B" = 2cos ("A" + "B")/2 * sin ("A" - "B")/2)]`
⇒ `(cos((theta - phi)/2))/(sin((theta - phi)/2))` = a
⇒ `cot((theta - phi)/2)` = a
⇒ `(theta - phi)/2 = cot^-1"a"`
⇒ θ – Φ = 2cot–1a
⇒ sin–1x – sin–1y = 2 cot–1a
Differentiating both sides w.r.t. x
`"d"/"dx" (sin^-1x) - "d"/"dx"(sin^-1x) = 2*"d"/"dx" cot^-1"a"`
⇒ `1/sqrt(1 - x^2) - 1/sqrt(1 - y^2) * "dy"/"dx"` = 0
⇒ `1/sqrt(1 - y^2) * "dy"/"dx" = 1/sqrt(1 - x^2)`
∴ `"dy"/"dx" = sqrt(1 - y^2)/sqrt(1 - x^2)`.
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