$$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$
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$\color{green}{\arctan\dfrac11=\arctan\dfrac12+\arctan\dfrac13.}$ – Lucian Dec 08 '14 at 03:16
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@Gunz, Have a look into http://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125/523626#523626 – lab bhattacharjee Dec 08 '14 at 05:07
3 Answers
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Let $x = \arctan \dfrac{1}{2}$ and $y = \arctan \dfrac{1}{3}$. Then, $\tan x = \dfrac{1}{2}$ and $\tan y = \dfrac{1}{3}$.
Using that formula, you can easily compute $\tan(x+y)$. Do you see how to get $x+y$ from that?
JimmyK4542
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Umm not really. Cause the equation uses arctan, while the hint use just tan – Gunz Dec 07 '14 at 23:48
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Using the formula, you can get $\tan(x+y) = \text{something}$. How do you solve that equation to get $x+y$? – JimmyK4542 Dec 07 '14 at 23:49
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Let $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)=u $$
Take tangents on both sides using hint given.
$$ \dfrac{1/2 +1/3}{1- {\dfrac{1} {6}}} = tan(u), $$
$$ tan(u) =1$$
$$ u = \pi/4 $$
Narasimham
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Yep, sum of tangents identity is the way to go. Write $x=arctan\frac{1}{2}$ and $y=arctan\frac{1}{3}$.
Then $tan(arctan\frac{1}{2}+arctan\frac{1}{3})=\frac{tan(arctan\frac{1}{2})+tan(arctan\frac{1}{3})}{1-(tan(arctan\frac{1}{2}))(tan(arctan\frac{1}{3}))}$
This is equal to $\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$
Thus $tan(x+y)=1$ and $x+y=\boxed{\frac{\pi}{4}}$