Question

In a triangle ABC, if $\cot A=(x^3+x^2+x{)}^{1/2}\cot B=(x+x^{-1}+1{)}^{1/2}$ and $\cot C=(x^{-3}+x^{-2}+x^{-1}{)}^{-1/2}$, then the triangle is

• A. equilateral
• B. isosceles
• C. right angled
• D. obtuse angled

Answer 1

The correct option is C right angled

$\cot (A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}=\frac{\sqrt{x^3+x^2+x}\sqrt{x+x^{-1}+1}-1}{\sqrt{x^3+x^2+x}+\sqrt{x+x^{-1}+1}}=\frac{x\sqrt{{(x+x^{-1}+1)}^2}-1}{(x+1)\sqrt{x+x^{-1}+1}}=\frac{x^2+1+x-1}{(x+1)\sqrt{x+x^{-1}+1}}=\frac{x}{\sqrt{x+x^{-1}+1}}=\frac{1}{\sqrt{x^{-3}+x^{-2}+x^{-1}}}=\cot C$
$\Rightarrow \cot (A+B)=\cot C\Rightarrow A+B=C\Rightarrow A+B-C=0$
And in triangle $A+B+C=\pi$
Therefore, we get $A+B=C=\frac{\pi }{2}$