Question

dy by dx of y=log √1-cosx by 1+cosx

Answer 1

  y  =  log [  (√1- cos x)  / (√1- +cos x)  ]
 1 - cos x = 2 sin² x/2      1+ cos x  = 2 cos² x/2
y  = log  sin x/2  / cos x/2 )  = log tan (x/2)
dy/dx =  1 / (tan x/2 )  *  sec²  x/2    * 1/2
           =  1 / [ 2 sin x/2 cos x/2 ]  = 1/sinx  = cosec x

Answer 2

y = log ( √1-cosx / √1+cosx )
y = log [ √1-(1-2sin²x/2) / √1+(2cos²x/2-1) ]
y = log [ √sin²x/2 / √cos²x/2 ]
y = log [√tan²x/2 ]
y = log [ tanx/2 ]
dy/dx = 1/ tanx/2 × sec²x/2.1/2
 on simplification u will get
dy/dx = 1 / 2. sinx/2.cosx/2
dy/dx = 1/sinx
dy/dx = cosecx