In triangle ABC,if r is the inradius and r_1, | vedclass

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(D) We know that $\sin 2C = 2 \sin C \cos C$ and $\sin 2B = 2 \sin B \cos B$. Substituting these into the expression: $\frac{1}{4}[b^2(2 \sin C \cos C

Vedclass · Accepted Answer

$rr_1 \cot \frac{A}{2}$

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(D) We know that $\sin 2C = 2 \sin C \cos C$ and $\sin 2B = 2 \sin B \cos B$. Substituting these into the expression: $\frac{1}{4}[b^2(2 \sin C \cos C) + c^2(2 \sin B \cos B)] = \frac{1}{2}[b^2 \sin C \cos C + c^2 \sin B \cos B]$. Using the sine rule,$\sin C = \frac{c \sin A}{a}$ and $\sin B = \frac{b \sin A}{a}$,or simply using the projection formula $a = b \cos C + c \cos B$ and $\Delta = \frac{1}{2} bc \sin A$: $\frac{1}{2}[b^2 \cos C \frac{2\Delta}{ab} + c^2 \cos B \frac{2\Delta}{ac}] = \frac{\Delta}{a}[b \cos C + c \cos B] = \frac{\Delta}{a}(a) = \Delta$. Now,consider the expression $rr_1 \cot \frac{A}{2}$. We know $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,and $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} = \frac{s(s-a)}{\Delta}$. Thus,$rr_1 \cot \frac{A}{2} = \frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{s(s-a)}{\Delta} = \Delta$. Therefore,$\frac{1}{4}[b^2 \sin 2C + c^2 \sin 2B] = rr_1 \cot \frac{A}{2}$.

In $\triangle ABC$,if $r$ is the inradius and $r_1, r_2, r_3$ are the ex-radii,then $\frac{1}{4}[b^2 \sin 2C + c^2 \sin 2B] =$

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