1.

In any triangle `A B C ,`prove that: `(b-c)/(b+c)=(tan((b-C)/2))/(tan((B+C)/2))`

Answer» By the sine rule, we have
`a/("sin A")=b/("sin B")=c/("sin C")="k (sky)"`
`rArr" a=ksin A, b=k sin B and c = ksinC"`.
`:." LHS"=((b-c))/((b+c))=("k sin B-k sin C")/("k sin B+k sin C")=("k(sin B - sin C)")/("k(sin B + sin C)")`
`=("(sin B - sin C)")/("(sin B + sin C)")=("2cos"((B+C))/2"sin"((B-C))/2)/("2sin"((B+C))/2"cos"((B-C))/2)`
`=("tan"1/2(B-C))/("tan"1/2(B+C))="RHS"`.
`"Hence,"((b-c))/((b+c))=("tan"1/2(B-C))/("tan"1/2(B+C))`


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