Question

$\frac{bc\cos A+ca\cos B+ab\cos C}{\cot A+\cot B+\cot C}=$

• A. $\Delta$
• B. $2\Delta$
• C. $\frac{1}{2}\Delta$
• D. ${\Delta }^2$

Answer 1

The correct option is B $2\Delta$

We know, $cotA=\frac{b^2+c^2-a^2}{4\Delta },cotB=\frac{a^2+c^2-b^2}{4\Delta },cotC=\frac{a^2+b^2-c^2}{4\Delta }$

$b.ccosA=\frac{b^2+c^2-a^2}{2},cosB.ca=\frac{a^2+c^2-b^2}{2},ab.cosC=\frac{a^2+b^2-c^2}{2}$

$\frac{bc\cos A+ca\cos B+ab\cos C}{\cot A+\cot B+\cot C}=$

$\frac{\frac{b^2+c^2-a^2}{2}+\frac{a^2+c^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}}{\frac{b^2+c^2-a^2}{4\Delta }+\frac{a^2+c^2-b^2}{4\Delta }+\frac{a^2+b^2-c^2}{4\Delta }}$

$=2\Delta$