Question

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

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

Answer 1

Answer: (A), (B)

The correct options are
A isosceles
B obtuse angled

Given $\cot A=(x^3+x^2+x{)}^{\frac{1}{2}}=x^{\frac{1}{2}}{(x^2+x+1)}^{\frac{1}{2}}$
$\cot B=(x+x^{-1}+1{)}^{\frac{1}{2}}={\left(\frac{x^2+x+1}{x}\right)}^{\frac{1}{2}}$
$\cot C=(x^{-3}+x^{-2}+x^{-1}{)}^{-\frac{1}{2}}={\left(\frac{x^3+x^2+1}{x^3}\right)}^{-\frac{1}{2}}={\left(\frac{x^3}{x^3+x^2+1}\right)}^{\frac{1}{2}}$
In triangle ABC,
$A+B+C=\pi$
$A+B=\pi -C$
$\tan (A+B)=-\tan C$
$\frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$
$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$
Converting in terms of cot, we get
$\cot B\cot C+\cot A\cot C+\cot A\cot B=1$
Substituting the given values,
${\left(\frac{x^2+x+1}{x}\right)}^{\frac{1}{2}}{\left(\frac{x^3}{x^3+x^2+1}\right)}^{\frac{1}{2}}+x^{\frac{1}{2}}{(x^2+x+1)}^{\frac{1}{2}}{\left(\frac{x^3}{x^3+x^2+1}\right)}^{\frac{1}{2}}+x^{\frac{1}{2}}{(x^2+x+1)}^{\frac{1}{2}}{\left(\frac{x^2+x+1}{x}\right)}^{\frac{1}{2}}=1$
$x{\left(\frac{x^2+x+1}{x^3+x^2+1}\right)}^{\frac{1}{2}}+x^2{\left(\frac{x^2+x+1}{x^3+x^2+1}\right)}^{\frac{1}{2}}+(x^2+x+1)=1$
$x{\left(\frac{x^2+x+1}{x^3+x^2+1}\right)}^{\frac{1}{2}}(1+x)=-(x^2+x)$
$[{\left(\frac{x^2+x+1}{x^3+x^2+1}\right)}^{\frac{1}{2}}+1](x^2+x)=0$
$\Rightarrow {\left(\frac{x^2+x+1}{x^3+x^2+1}\right)}^{\frac{1}{2}}=-1;x=0,x=-1$
$\Rightarrow x^3-x=0$
$\Rightarrow x=0,1,-1$
Only $x=1$ can be taken, else it will give complex values
$\therefore \cot A=\sqrt{3},\cot B=\sqrt{3}$ and $\cot C=\frac{1}{\sqrt{3}}$
So, the triangle is isosceles