How to simplify $$\arctan \left(\frac{1}{2}\tan (2A)\right) + \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) $$ for $0< A< \pi /4$?

This is one of the problems in a book I'm using. It is actually an objective question , with 4 options given , so i just put $A=\pi /4$ (even though technically its disallowed as $0< A< \pi /4$) and got the answer as $\pi $ which was one of the options , so that must be the answer (and it is weirdly written in options as $4 \arctan (1) $ ).

Still , I'm not able to actually solve this problem. I know the formula for sum of three arctans , but it gets just too messy and looks hard to simplify and it is not obvious that the answer will be constant for all $0< A< \pi /4$. And I don't know of any other way to approach such problems.


As $0<A<\dfrac\pi4\implies\cot A>1\implies\cot^3A>1$

Like showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$,

$\arctan(\cot A)+\arctan(\cot^3A)=\pi+\arctan\left(\dfrac{\cot A+\cot^3A}{1-\cot A\cdot\cot^3A}\right)$

Now $\dfrac{\cot A+\cot^3A}{1-\cot A\cdot\cot^3A}=\dfrac{\tan^3A+\tan A}{\tan^4A-1}=\dfrac{\tan A}{\tan^2A-1}=-\dfrac{\tan2A}2$

and $\arctan(-x)=-\arctan(x)$

  • $\begingroup$ Nice , thank you! the mistake i did was using formula for summing all three at the same time. $\endgroup$ – A Googler Jan 22 '15 at 7:08

Over the given interval we have $\arctan\cot A=\frac{\pi}{2}-A$ and, by setting $t=\tan A$: $$\begin{eqnarray*}&&\tan\left(\arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)\right)=\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}\\&=&\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}=\frac{\frac{1}{t^3}+\frac{t}{1-t^2}}{1-\frac{1}{t^2(1-t^2)}}=\frac{1-t^2+t^4}{t(t^2-t^4-1)}=-\frac{1}{t}\end{eqnarray*}$$ so: $$ \arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)=\frac{\pi}{2}+A $$ and the sum of the three arctangents is $\color{red}{\pi}$ as wanted. Another chance is given by differentiating such a sum wrt to $A$ and check that the derivative is zero oven the given interval, so the sum equals its value in the point $A=\frac{\pi}{8}$, for instance.

  • $\begingroup$ Thank you! I didn't recognise that arctan cot a is pi/2 -a , also that t trick was good. $\endgroup$ – A Googler Jan 22 '15 at 7:09
  • $\begingroup$ Since tan(stuff)=-1/tanA , wouldn't we get tan(stuff)=-cotA=-tan(pi/2-A)=tan(A-pi/2) so that sum of arctans is A-pi/2? What have i done wrong? $\endgroup$ – A Googler Jan 22 '15 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.