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In any `triangle ABC`, prove that than `((A-B)/(2))=((a-b)/(a+b))cot .(C)/(2)`. |
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Answer» We know that, `(a)/(sinA)=(b)/(sinB)=K` `a=KsinA,b=KsinB` `(a-b)/(a+b)=(KsinA-Ksinb)/(KsinA+KsinB)` `=(sinA-sinB)/(sinA+sinB)` `=(2"cos"(A+B)/(2)"sin"(A-B)/(2))/(2"sin"(A+B)/(2)"cos"(A-B)/(2))` `(a-b)/(a+b)=("tan"(A-B)/(2))/("tan"(A+B)/(2))` `="cot"(A+B)/(2)"tan"(A-B)/(2)` We know that, `(A+B+C)/(2)=90^(@),(A+B)/(2)=90^(@)-(C)/(2)` `cot((A+B)/(2))=cot(90^(@)-(C)/(2))` `="tan"(C)/(2)` `(a-b)/(a+b)="tan"(C)/(2)"tan"(A-B)/(2)` `"tan"(A-B)/(2)=((a-b)/(a+b))"cot"(C)/(2)` |
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