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The identity is:
atan(x) + acot(x) = pi/2
Your equation is
atan(x) + acot(1/3) = pi/2
The only answer is x = 1/3
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?
To solve for x, we proceed as below:
Taking tangent of the angles on both sides we get:
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:
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
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