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

Answer 1

Answer: (C)

right angled

Given, $\cot A=(x^3+x^2+x{)}^{\frac{1}{2}}$.

And

$\cot C=(x^{-3}+x^{-2}+x^{-1}{)}^{-\frac{1}{2}}$

Hence

$\tan C=(x^{-3}+x^{-2}+x^{-1}{)}^{\frac{1}{2}}$

$=\frac{\sqrt{1+x+x^2}}{x\sqrt{x}}$

$=\frac{\sqrt{x}cotB}{x\sqrt{x}}$

$x\tan C=\cot B$

$\cot C.\cot B=x$

If $x=1$

We get, $\cot C\cot B=1$

Then

$B=\frac{\pi }{2}-C$

Hence a right angled triangle.