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प्रश्न
`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`
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उत्तर
Let y = `tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2)))`
Putting x2 = cos 2θ
∴ θ = `1/2 cos^-1 x^2`
y = `tan^-1 ((sqrt(1 + cos 2theta) + sqrt(1 - cos 2theta))/(sqrt(1 + cos 2theta) - sqrt(1 - cos 2theta)))`
⇒ y = `tan^-1 ((sqrt(2cos^2theta) + sqrt(2sin^2theta))/(sqrt(2cos^2theta) - sqrt(2sin^2theta)))`
⇒ y = `tan ((sqrt(2) cos theta + sqrt(2) sin theta)/(sqrt(2) cos theta - sqrt(2) sin theta))`
⇒ y = `tan^-1 ((cos theta + sin theta)/(cos theta - sin theta))`
⇒ y = `tan^-1 [((costheta)/(costheta) + (sintheta)/(costheta))/((costheta)/(costheta) - (sintheta)/(costheta))]`
⇒ y = `tan^-1 [(1 + tan theta)/(1 - tan theta)]`
⇒ y = `tan^-1 [(tan pi/4 + tan theta)/(1 - tan pi/4 * tan theta)]`
⇒ y = `tan^-1 [tan (pi/4 + theta)]`
⇒ y = `pi/4 + theta`
⇒ y = `pi/4 + 1/2 cos^-1 x^2`
Differentiating both sides w.r.t. x
`"dy"/"dx" = "d"/"dx" (pi/4) + 1/2 "d"/"dx" (cos^-1 x^2)`
= `0 + 1/2 xx (-1)/sqrt(1 - x^4) * "d"/"dx" (x^2)`
= `(-1.2x)/(2sqrt(1 - x^4)`
= `- x/sqrt(1 - 4x^4)`
Hence, `"dy"/"dx" = - x/sqrt(1 - x^4)`.
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