Question

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"` = ...........

पर्याय

• `x/sqrt(1 - x^2)`

• `(1 - 2x)/sqrt(1 - x^2)`

• `(1 - 2x)/(2sqrt(1 - x^2)`

• `(1 - 2x^2)/sqrt(1 - x^2)`

Answer 1

`(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))`.