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In any triangle `A B C ,`prove that: `("sin"(B-C))/("sin"(B+C))=(b^2-c^2)/(a^2)` |
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Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")="k(say)"` `rArr" a=ksin A, b=ksin B and c=ksin C"`. `:." RHS"=((b^(2)-c^(2)))/a^(2)=(k^(2)sin^(2)B-k^(2)sin^(2)C)/(k^(2)sin^(2)A)` `=((sin^(2)B-sin^(2)C))/(sin^(2)A)=(sin(B+C)" sin "(B-C))/(sin^(2)(B+C))` `[{:(because,(A+B+C=pirArrA=pi-(B+C))),(therefore,sin A =sin{pi-(B+C)}=sin(B+C)):}]` `=("sin (B-C)")/("sin (B+C)")="LHS"`. `:." RHS=LHS"`. `"Hence, "("sin (B-C)")/("sin (B+C)")=((b^(2)-c^(2)))/a^(2)` |
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