Evaluating $\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{8}\right)$

Question

Find the exact value of $$\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{8}\right)$$

How do you solve this problem? I don't know what trigonometric properties or identities may help with it.

Answer 1

If $$x=\arctan\frac 12+\arctan\frac 15+\arctan\frac 18$$ you can take the tangent of both sides: $$\tan x=\tan\left(\arctan\frac 12+\arctan\frac 15+\arctan\frac 18\right)$$ Then you will need to find a formula for $\tan(a+b+c)$. I don't remember such formula, but you can write it as $\tan(a+(b+c))$. So you need a formula for $\tan(a+b)$. The answer for this step is $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ If you don't remember this formula, you can derive it from $$\tan(a+b)=\frac{\sin(a+b)}{\cos(a+b)}$$ The rest is simple arithmetic. At the end you need to check if $\arctan \tan x$ is the right solution, or is it shifted by some multiples of $\pi$

Answer 2

Let $a,b,c$ be angles, so that $s=\tan a$, $t=\tan b$, $u=\tan c$ make sense. The formula $$ \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b} =\frac {s+t}{1-st} $$ extends easily by using it twice (write $a+b+c=a+(b+c)$) to $$ \tan(a+b+c)= \frac{s+\frac{t+u}{1-tu}}{1-s\cdot\frac{t+u}{1-tu}} =\frac{s+t+u-stu}{1-st-tu-ut}\ . $$ In our case, $s,t,u$ are $\frac 12$, $\frac 15$, $\frac 18$, so the tangent of the sum to be computed is $$ \frac {\frac 12+\frac 15+\frac 18-\frac 1{2\cdot 5\cdot 8}} {1-\frac 1{2\cdot 5}-\frac 1{5\cdot 8}-\frac 1{8\cdot 2}} =\frac {2\cdot 5+5\cdot 8+8\cdot 2-1}{2\cdot 5\cdot 8-2-5-8} =\frac{65}{65}=1\ . $$ Observing that the sum to be computed is positive and $<3\pi/4$, we get the value $\boxed{\pi/4}$ for it.