Question

Tg(arccos1-2arctg1/корень из 3)

Answer 1

Учтём, что arcCos1 = 0, arc tg1/√3 = π/6
наш пример:
tg (-2π/6) = -tg π/3 = -√3

Answer 2

To simplify the expression Tg(arccos(1 - 2arctg(1/√3))), let's break it down step by step.

First, let's consider the innermost function: arctg(1/√3). The tangent of an angle is equal to the ratio of the opposite side to the adjacent side in a right triangle. In this case, we have a right triangle with an opposite side of 1 and an adjacent side of √3.

Using the Pythagorean theorem, we can find the hypotenuse of the triangle:

h = √(1^2 + (√3)^2) = √(1 + 3) = 2

So, the tangent of the angle is 1/√3. We can find the angle itself using the inverse tangent function:

arctg(1/√3) = arctg(√3/3)

Next, we substitute this result into the expression Tg(arccos(1 - 2arctg(√3/3))). Now, let's consider the arccos function. The cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse in a right triangle. In this case, we can form a right triangle with an adjacent side of (1 - 2arctg(√3/3)) and a hypotenuse of 1.

Using the Pythagorean theorem, we can find the opposite side of the triangle:

opposite = √(1^2 - (1 - 2arctg(√3/3))^2)

Now, we substitute this into the expression Tg(arccos(1 - 2arctg(√3/3))):

Tg(arccos(1 - 2arctg(√3/3))) = Tg(arccos(1 - 2arctg(√3/3))) = Tg(arccos(1 - 2arctg(√3/3)))

At this point, the expression cannot be simplified further without more specific values for the angles involved. If you provide specific values for the angles, I can calculate the result for you.