Question

If 2y = `(cot^-1((sqrt3cosx + sinx)/(cosx - sqrt3 sinx)))^2`, x ∈ `(0, pi/2)` then `dy/dx` is equal to ______

पर्याय

• `pi/6 - x`

• `x - pi/6`

• `pi/3 - x`

• `2x - pi/3`

Answer 1

If 2y = `(cot^-1((sqrt3cosx + sinx)/(cosx - sqrt3 sinx)))^2`, x ∈ `(0, pi/2)` then `dy/dx` is equal to `underline(x - pi/6)`.

Explanation:

2y = `(cot^-1((sqrt3cosx + sinx)/(cosx - sqrt3 sinx)))^2`

= `(cot^-1((sqrt3 + tanx)/(1 - sqrt3tanx)))^2`

= `(cot^-1((tan  pi/3 + tanx)/(1 - tan  pi/3tanx)))^2`

= `{cot^-1[tan(pi/3 + x)]}^2`

= `{pi/2 - tan^-1[tan(pi/3 + x)]}^2`

= `{:{((pi/2 - pi/3 - x)^2","   0 < x < pi/6), ((pi/2 + (2pi)/3 - x)^2","   pi/6 < x < pi/2):}`

∴ 2y = `{((pi/6 - x)^2","   0 < x < pi/6), (((7pi)/6 - x)^2","  pi/6 < x < pi/2):}`

∴ `dy/dx = {(x -  pi/6","   0 < x < pi/6), (x - (7pi)/6","   pi/6 < x < pi/2):}`