You can answer this question by recalling (or looking up) an obscure trigonometric identity (see the link):

tan(x) + atan(1/x) = pi/2 if x > 0

. = -pi/2 if x < 0

From the second term (1/x = 1/3), we know that x = 3

To prove this identity, we can make use of well known pi/2 phase shift identities of trig functions, in this case tan(y) = cot(pi/2 - y)

Let y = atan(x); then x = tan(y) = cot(pi/2 - y)

x = cot(pi/2 - y)

acot(x) = acot(cot(pi/2 - y)) = pi/2 - y = pi/2 - atan(x)

--> pi/2 = atan(x) + acot(x)

This form is more well known than the first form given above. To get to that form, you have to recall that,

acot(x) = atan(1/x) , x > 0

acot(x) = -pi + atan(1/x) , x < 0

It`s arcctg (1/3) .. and as i know arcctg (x)=y and ctg (y)=x .

So if arcctg(1/3)=x =>ctg (x)=1/3 . And i don`t know any value of x to solve this equation.

It can be the exercise wrong?

arctgx+arctg (1/3)=pi/2

To solve for x, we proceed as below:

Taking tangent of the angles on both sides we get:

tan(arctanx+arctan1/3)=tanpi/2

[x+1/3]/(1-x*(1/3)]= +or-inf

Threfore, denominator =0.

Or 1-x/3=0 or x=3.

The solution can also be got from a right angled triangle, ABC in which B is right angle:

Given arctgx+arctg(1/3)=Pi/2

In a right angled triangle with B =Pi/2, angle A+ angle C=Pi/2

Let tan C = 1/3 = BC/AB, then , arctg (1/3 )= C by definition.

Therefore Arctg x = arc tan A =arctg(AB/BC)= arctg(3) or x= 3