Question

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

Options

• `π/6 - x`

• `π/3 - x`

• `x - π/6`

• `2x - π/3`

Answer 1

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

Explanation:

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

2y = `(cot^-1((sqrt(3)/2 cosx + 1/2sinx)/(1/2cosx - sqrt(3)/2sinx)))^2` 

2y = `(cot^-1((sin(π/3 + x))/(cos(π/3 + x))))^2`

= `(cot^-1tan(π/3 + x))^2`

= `(π/2 - tan^-1tan(π/3 + x))^2`

= 2y = `{{:((π/2 - (π/3 + x))^2, 0 < x < x/6),((π/2 - tan^-1 tan(π - (π/3 + x)))^2, π/6 < x < π/2):}`

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

`(2dy)/(dx) = {{:(2(π/2 - x)(-1),  0 < x < π/6),(2((7π)/6 - x)(-1),  π/6 < x < (7π)/6):}`

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