Question

In a ∆ ABC ,if c-a = mb then;
a.cot(A-B)/2 = (1+mcosA)/msinA
b.cot(B-C)/2 = (1+mcosB)/msinB
c.cot(C-A)/2 = (1+mcosC)/msinC
d.cot(C-A)/2 = (1+mcosA)/msinA​

Answer 1

Answer:

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