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Hii !!
We have :-
LHS = Tan^-1 ( √1 - Cos x / 1 + Cos x ) = Tan^-1 √2 Sin² ( x/2 ) / 2 Cos² ( x/2 ).
=> Tan^-1 ( Tan x/2 ) = x/2 = RHS.
Therefore,
Tan^-1 ( √1 - Cos x / 1 + Cosx ) = x/2.
We have :-
LHS = Tan^-1 ( √1 - Cos x / 1 + Cos x ) = Tan^-1 √2 Sin² ( x/2 ) / 2 Cos² ( x/2 ).
=> Tan^-1 ( Tan x/2 ) = x/2 = RHS.
Therefore,
Tan^-1 ( √1 - Cos x / 1 + Cosx ) = x/2.
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Hi!!!!!
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