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प्रश्न
Differentiate `tan^-1 (sqrt(1 - x^2)/x)` with respect to`cos^-1(2xsqrt(1 - x^2))`, where `x ∈ (1/sqrt(2), 1)`
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उत्तर
Let u = `tan^-1 (sqrt(1 - x^2)/x)` and v = `cos^-1(2xsqrt(1 - x^2))`.
We want to find `"du"/"dv" = (("du")/("dx"))/(("dv")/("dx"))`
Now u = `tan^-1 (sqrt(1 - x^2)/x)`.
Put x = `sintheta. (pi/2 < theta < pi/2)`
Then u = `tan^-1 (sqrt(1 - sin^2theta)/sintheta)`
= `tan^-1 (cot theta)`
= `tan^-1 {tan (pi/2 - theta)}`
= `pi/2 - theta`
= `pi/2 - sin^-1x`
Hence `"du"/"dx" = (-1)/sqrt(1 - x^2)`.
Now v = `cos^-1 (2x sqrt(1 - x^2))`
= `pi/2 - sin^-1 (2x sqrt(1 - x^2))`
= `pi/2 - sin^-1 (2sintheta sqrt(1 - sin^2theta))`
= `pi/2 - sin^-1 (sin 2theta)`
= `pi/2 - sin^-1 {sin (pi - 2theta)}` .......{Since `pi/2` < 2θ < π]
= `pi/2 - (pi / 2theta)`
= `(-pi)/2 + 2theta`
⇒ v = `(-pi)/2 + 2sin^-1x`
⇒ `"dv"/"dv" = (("du")/("d"x))/(("dv")/("dx"))`
= `((-1)/sqrt(1 - x^2))/(2/sqrt(1 - x^2))`
= `(-1)/2`
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