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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

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

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