Mathematics - Properties of Triangle Question with Solution | TestHub

Given that $a^2,b^2,c^2$ are in A.P., we have $2b^2=a^2+c^2$. We know the formula for cotangent of an angle in a triangle: $\cot A=\frac{b^2+c^2-a^2}{4\Delta }$, $\cot B=\frac{a^2+c^2-b^2}{4\Delta }$, $\cot C=\frac{a^2+b^2-c^2}{4\Delta }$. For $\cot A,\cot B,\cot C$ to be in A.P., we must have $2\cot B=\cot A+\cot C$. Substituting the expressions: $2\frac{a^2+c^2-b^2}{4\Delta }=\frac{b^2+c^2-a^2}{4\Delta }+\frac{a^2+b^2-c^2}{4\Delta }$. Multiplying by $4\Delta$: $2(a^2+c^2-b^2)=(b^2+c^2-a^2)+(a^2+b^2-c^2)$. This simplifies to $2a^2+2c^2-2b^2=2b^2$, which means $2a^2+2c^2=4b^2$, or $a^2+c^2=2b^2$. This is exactly the given condition. Hence, $\cot A,\cot B,\cot C$ are in A.P.