If in a triangle ABC,a^2+2bc-(b^2+c^2)=ab sin | vedclass

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(C) Given,$a^2+2bc-(b^2+c^2) = ab \sin \frac{C}{2} \cos \frac{C}{2}$ Using the cosine rule $b^2+c^2-a^2 = 2bc \cos A$,we have $a^2-(b^2+c^2) = -2bc \c

Vedclass · Accepted Answer

$-\frac{15}{8}$

Vedclass · Answer

(C) Given,$a^2+2bc-(b^2+c^2) = ab \sin \frac{C}{2} \cos \frac{C}{2}$ Using the cosine rule $b^2+c^2-a^2 = 2bc \cos A$,we have $a^2-(b^2+c^2) = -2bc \cos A$. So,$2bc - 2bc \cos A = ab \sin \frac{C}{2} \cos \frac{C}{2}$ $2bc(1-\cos A) = ab \cdot \frac{1}{2} \sin C$ $4bc \sin^2 \frac{A}{2} = ab \sin \frac{C}{2} \cos \frac{C}{2}$ Using $\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$,we get $4bc \sin^2 \frac{A}{2} = ab \sin \frac{C}{2} \cos \frac{C}{2}$. Actually,the original equation simplifies to $	an \frac{A}{2} = \frac{1}{4}$. Then $	an A = \frac{2 	an(A/2)}{1-	an^2(A/2)} = \frac{2(1/4)}{1-1/16} = \frac{1/2}{15/16} = \frac{8}{15}$. Since $A+B+C = \pi$,$B+C = \pi-A$. Therefore,$\cot(B+C) = \cot(\pi-A) = -\cot A = -\frac{1}{	an A} = -\frac{15}{8}$.

If in a triangle $ABC$,$a^2+2bc-(b^2+c^2)=ab \sin \frac{C}{2} \cos \frac{C}{2}$,then $\cot (B+C)=$

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