If `x = tan""(y)/(2) - log[((1 + tan""(y)/(2))^(2))/(tan""(y)/(2))], "then " dy/(dx)` is equal to

Options

A. ` (1)/(2) "" sin y (1 - sin y + cos y )`
B. `sin y (1 - sin y - cos y)`
C. `(1)/(2) "" sin y (1 + sin y + cos y)`
D. None of the above

Answer

C

Explanation

`x = tan(y)/(2) - [((1 + tan""(y)/(2))^(2))/("tan" (y)/(2))]` ...(i)
Put ` tan"" (y)/(2) = t rArr (y)/(2) = tan^(-1) t `
` rArr (dy)/(dx) = (2)/(1 + t^(2)) = (2) /(1 + tan^(3) ((y)/(2)) = 2 cos^(2) ((y)/(2)) = 1 + cos ` ….(ii)
From Eq . (i) , ` x = t = " In" [ ((1 + t)^(2))/(t)]`
`(dy)/(dx) = 1 - (t)/((1 + t)^(2)) ((1)/(t^(2)) + 1) = (t^(2) +1)/( t^(2) +t)`
`[ because (d)/(dt) ((1 + t)^(2))/(t) = d/dt ((1)/(t) + 2 + t) = - (1)/(t^(2)) + 1]`
` (dx)/(dt) = (1 + tan^(2)""(y)/(2))/(1 + tan^(2)"tan"(y)/(2))= (2)/(1 - cos y + siny) ` ...(iii)
On dividing Eq. (ii) by .(iii) , we get
`rArr (dy)/(dx) = ((1 + cos y) (1 - cos y + sin y))/(2)`
` dy/dx = 1/2"sin" y (1 + sin y + cos y)` .