Question

If y=log [square root of (1-cos x /1+cos x)], find dy/dx.

Answer 1

y= log [sq.root of (1-cosx / 1+cosx)]

1-cosx   x 1-cosx   =   (1-cosx)^2
1+cosx      1+cosx         1 - cos^2x
                              =   (1-cosx)^2
                                     sin^2x
on taking square root we get
                             = 1- cosx  = cosecx - cot x
                                    sinx
now, y = log(cosecx - cotx)
       dy/dx =   1 / (cosecx - cotx) x (-cosec^2x - logsinx)
                =  (-cosec^2x - logsinx) /  (cosecx - cotx)
note: differentiate part alone doubt whether it is right or wrong