Question

3arctg(-корень из 3/3) + arcctg(tg0)=?

Answer 1

Arctg√3=пи/3
tgпи/3=tg60=√3

Answer 2

To simplify the expression, let's first find the values of the trigonometric functions involved:

1. arctg(-√3/3): The tangent function is negative in the second and fourth quadrants. In the second quadrant, both the sine and cosine functions are positive. Let's consider a right triangle with an angle θ in the second quadrant, where the opposite side is √3 and the adjacent side is 3. Then:

   tg(θ) = -√3/3

   We can find the hypotenuse using the Pythagorean theorem:

   hypotenuse^2 = (√3)^2 + 3^2 hypotenuse^2 = 3 + 9 hypotenuse^2 = 12 hypotenuse = √12 = 2√3

   Now, we can find the value of arctg(-√3/3):

   arctg(-√3/3) = θ

   So, θ is the angle in the second quadrant whose tangent is -√3/3.

2. arcctg(tg(0)): The cotangent function is the reciprocal of the tangent function. Since tg(0) = 0, we have:

   ctg(0) = 1/tg(0) = 1/0

   However, division by zero is undefined, and the arcctg function is not defined for ctg(0) since ctg(0) is undefined.

So, the expression 3arctg(-√3/3) + arcctg(tg0) is undefined due to the division by zero when trying to find the arcctg(tg(0)) term.