If `2y=(cot^(-1)((sqrt3 cos x + sin x)/(cos x - sqrt3 sin x)))^(2) , x in (0, pi/2) " then " (dy)/(dx) ` is equal to

Options

A. `pi/6 - x`
B. ` x- pi/6`
C. ` pi/3 - x`
D. ` 2x - pi/3`

Answer

B

Explanation

Given expression is
`2y=(cot^(-1)((sqrt3 cos x + sin x)/( cos x -sqrt3 sin x)))^(2) = (cot^(-1)((sqrt3 cot x +1)/(cot x - sqrt3)))^(2)`
[dividing each term of numerator and denominator by sin x]
`=(cot^(-1)((cot. pi/6 cot x + 1)/(cot x - cot .pi/6)))^(2)" "[:' cot. pi/6 =sqrt3]`
`=(cot^(-1)(cot(pi/6 - x)))^(2)" "[:' cot (A-B)=(cot A cot B + 1)/(cot B - cot A) ]`
`={{:(" "(pi/6 - x)^(2)", " 0 lt x lt pi/6),((pi+(pi/6 - x))^(2)", " pi/6 lt x lt pi/2):}" "[:' cot^(-1) (cot theta) ={{:(pi+theta", " -pi lt theta lt 0),(theta", " 0 lt thetalt pi),(theta-pi", " pi lt thetalt 2pi):}]`
`rArr 2y={{:((pi/6 - x)^(2)", " 0 lt x lt pi/6),(((7pi)/6-x)^(2)", " pi/6 lt x lt pi/2):}`
`rArr 2(dy)/(dx) ={{:(2(pi/6-x)(-1)", " 0 lt xlt pi/6),(2((7pi)/6 - x) (-1)", " pi/6 lt xlt pi/2):}`
`rArr (dy)/(dx) ={{:(x-pi/6", " 0 lt xlt pi/6),(x-(7pi)/6", " pi/6 lt xlt pi/2):}`