Question

Ctg(6arctg1/√3-arcctg(-√3/3))

Answer 1

Ctg(6*п/6-п-п/3)=ctg(-п/3)=- $\sqrt3$/3

Answer 2

Ctg(п-п-п/3)=ctg(-п/3)=-корень3на3

Answer 3

To find the value of the expression $\cot\left(6\arctan\left(\frac{1}{\sqrt{3}}\right) - \text{arccot}\left(-\frac{\sqrt{3}}{3}\right)\right)$, we can use trigonometric identities and simplify the expression step by step.

1. First, let's simplify the angles inside the trigonometric functions: $\arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ (This is a common value for $\frac{1}{\sqrt{3}}$ in the context of trigonometry.)

   $\text{arccot}\left(-\frac{\sqrt{3}}{3}\right) = \frac{5\pi}{6}$ (The cotangent of an angle is the reciprocal of the tangent, so the cotangent of $\frac{\pi}{6}$ is $\frac{1}{\tan(\frac{\pi}{6})}$, which is $\frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$. Thus, the angle whose cotangent is $-\sqrt{3}$ is $\frac{5\pi}{6}$.)

2. Now, we can rewrite the expression with these simplified angles: $\cot\left(6\left(\frac{\pi}{6}\right) - \frac{5\pi}{6}\right)$

3. Simplify the inside of the cotangent function: $6\left(\frac{\pi}{6}\right) - \frac{5\pi}{6} = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$

4. Now, we have the expression $\cot\left(\frac{\pi}{6}\right)$.

5. The cotangent of $\frac{\pi}{6}$ is $\sqrt{3}$.

So, $\cot\left(6\arctan\left(\frac{1}{\sqrt{3}}\right) - \text{arccot}\left(-\frac{\sqrt{3}}{3}\right)\right) = \sqrt{3}$.