# Arctan identity

#### moshelevi33

Hey I'm trying to answer this without a calculator.
Arctan(2+sqrt(3)) with calculator it is 75.

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#### Compendium

I assume that is in degrees. You can reference a table of the unit circle to see that is the case. If you don't have a table that precise, use the half-angle formula of tangent for 150 degrees.

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#### skipjack

Forum Staff
tan(75Â°) = tan(45Â° + 30Â°) = (tan(45Â°) + tan(30Â°))/(1 - tan(45Â°)tan(30Â°)).

As tan(45Â°) = 1 and tan(30Â°)= 1/âˆš3, tan(75Â°) = (âˆš3 + 1)/(âˆš3 - 1) = 2 + âˆš3.

• 1 person

#### skeeter

Math Team
$0 < t < 90$

$\tan{t}=2+\sqrt{3} \implies \sin{t}=2\cos{t}+\sqrt{3}\cos{t}$

$2\cos{t}=\sin{t}-\sqrt{3}\cos{t}$

$\cos{t}=\dfrac{1}{2}\sin{t}-\dfrac{\sqrt{3}}{2}\cos{t}$

$\cos{t}=\sin(150)\sin{t}+\cos(150)\cos{t}$

$\cos{t}=\cos(150-t) \implies t=150-t \implies t = 75$

• 1 person

#### greg1313

Forum Staff
$$2+\sqrt3=\frac{\sin60+\sin90}{\sin150}= \frac{2\sin^275}{2\cos75\sin75}=\tan75\implies \arctan(2+ \sqrt3)=75^\circ$$

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• 1 person

#### skipjack

Forum Staff
Using cot(A) = csc(2A) + cot(2A), 2 + âˆš3 = csc(30Â°) + cot(30Â°) = cot(15Â°) = tan(75Â°).

• 1 person

#### topsquark

Math Team
Wow! The solutions just keep on coming! :dance:

-Dan

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