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प्रश्न
Choose the correct option from the given alternatives :
If y = `tan^-1(x/(1 + sqrt(1 - x^2))) + sin[2tan^-1(sqrt((1 - x)/(1 + x)))] "then" "dy"/"dx"` = ...........
विकल्प
`x/sqrt(1 - x^2)`
`(1 - 2x)/sqrt(1 - x^2)`
`(1 - 2x)/(2sqrt(1 - x^2)`
`(1 - 2x^2)/sqrt(1 - x^2)`
उत्तर
`(1 - 2x)/(2sqrt(1 - x^2)`
`y = tan^-1(x/(1 + sqrt(1 - x^2))) + sin[2tan^-1sqrt((1 - x)/(1 + x))]`
Put x = cos θ. Then θ = cos–1x
`∴ y = tan^-1((cosθ)/(1 + sqrt(1 - cos^2θ))) + sin[2tan^-1sqrt((1 - cosθ)/(1 + cosθ))]`
`= tan^-1((cosθ)/(1 + sinθ)) + sin[2tan^-1sqrt((2sin^2(θ/2))/(2cos^2(θ/2)))]`
`= tan^-1[(sin(pi/2 - θ))/(1 + cos(pi/2 - θ))] + sin[2tan^-1(tan θ/2)]`
`= tan^-1[(2sin(pi/4 - θ/2).cos(pi/4 - θ/2))/(2cos^2(pi/4 - θ/2))] + sin(2 xx θ/2)`
`= tan^-1[tan(pi/4 - θ/2) + sinθ]`
`= pi/4 - θ/2 + sqrt(1 - cos^2θ)`
= `pi/4 - 1/2cos^-1x + sqrt(1 - x^2)`
∴ `"dy"/"dx" = 0 - 1/2 xx (-1)/sqrt(1 - x^2) + (1)/(2sqrt(1 - x^2)) xx (-2x)`
= `(1)/(2sqrt(1 - x^2)) - x/sqrt(1 - x^2)`
= `(1 - 2x)/(2sqrt(1 - x^2))`.
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