Question

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

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

Answer 1

Answer: (C)

right angled

From above,
We have $\cot (A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A}$
$=\frac{{(x^3+x^2+x)}^{\frac{1}{2}}{(x+x^{-1}+1)}^{\frac{1}{2}}-1}{{(x^3+x^2+x)}^{\frac{1}{2}}+{(x+x^{-1}+1)}^{\frac{1}{2}}}$
$=\frac{x^{\frac{1}{2}}{(x^2+x+1)}^{\frac{1}{2}}.\frac{1}{x^{\frac{1}{2}}}{\left((x^2+x+1)\right)}^{\frac{1}{2}}-1}{x^{\frac{1}{2}}(x^2+x+1)+\frac{1}{x^{\frac{1}{2}}}(x^2+x+1)}$
$=\frac{{\left({(x^2+x+1)}^{\frac{1}{2}}\right)}^2-1}{{(x^2+x+1)}^{\frac{1}{2}}(\sqrt{x}+\frac{1}{\sqrt{x}})}$
$=\frac{x^2+x}{\frac{{(x^2+x+1)}^{\frac{1}{2}}(x+1)}{\sqrt{x}}}$
$=\frac{x(x+1)\sqrt{x}}{\sqrt{x^2+x+1}(x+1)}$
$=\frac{x\sqrt{x}}{\sqrt{x^2+x+1}}$
$=x^{\frac{3}{2}}{(x^2+x+1)}^{\frac{-1}{2}}$
$={(x^{-1}+x^{-2}+x^{-3})}^{\frac{1}{2}}$
$=\cot C$
$\Rightarrow A+B=C$ and $A+B+C=\pi$
$\Rightarrow C+C=\pi$
$\therefore C=\frac{\pi }{2}$
Hence, the triangle is right-angled triangle