# Prove that arctan (2+sqrt3)+arctan (2-sqrt3)=pi/2!

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### 3 Answers

Let us assume that:\

arctan (2+ sqrt3) = x

arcttan (2-sqrt3) = y

==> tanx = 2+ sqrt3

==> tany = 2-sqrt3

Now we know that:

tan (x+y) = (tanx + tany)/[1- tanx*tany]

==> tan (arctan (2+ sqrt3) + arctan(2-sqrt3) = tan (x+y)

= (2+sqrt + 2-sqrt3)/ (1- (2+sqrt3)(2-sqrt3)

= 4/ 1- 1

= 4/0= inf

==> tan[arcttan (2+sqrt3) + arctan (2-sqrt3) ]= infinity

==> arcttan( 2+sqrt3) + arctan (2-sqrt3) = pi/2

To find arctan(2+sqrt3)+arctan(2-sqrt3)

Solution:

We use arc tan a+arc tanb = arc tan (a+b)/(1-ab).

a = 2+sqrt2

b=2-sqrt3.

Therefore

(a+b)/(1-ab) = {2+sqrt3 +2-sqrt}/(1- (2+sqrt3)(2-sqrt3)}

= 4/{1- [2^2-(sqt3)^2]}

= 4/{1-(4-3)}

=4/0

Therefore

arc tan (2+sqrt3) + arc tan (2-sqrt3)} = 4/0 = inf.

But artan (inf) = pi/2 , as tan (pi/2 from left) = infinity.

The terms arctan (2+sqrt3) and arctan (2-sqrt3) are angles.

Let's note arctan (2+sqrt3) = x and arctan (2-sqrt3) = y

We'll apply the tangent function to the addition of angles x and y:

tan (x+y) = (tan x + tany)/(1 - tan x*tany)

where:

tan x = tan [arctan (2+sqrt3)] = 2+sqrt3

tan y = tan [arctan (2-sqrt3)] = 2-sqrt3

tan x*tan y = (2+sqrt3)(2-sqrt3) = 4-3 = 1

Let's substitute the values into the formula:

tan (x+y) = (2+sqrt3+2-sqrt3) / (1 - 1)

We'll eliminate like terms:

tan (x+y) = 4/0

tan (x+y) = infinite => x+y = pi/2

**arctan (2+sqrt3)+arctan (2-sqrt3)=pi/2 q.e.d.**