Review question

If $bc=a^2+1$, what is $\arctan[(a+b)^{-1}]+\arctan[(a+c)^{-1}]$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5129

Solution

The positive numbers $a$, $b$ and $c$ satisfy $bc=a^2+1$. Prove that $\arctan\left(\frac{1}{a+b}\right)+\arctan\left(\frac{1}{a+c}\right)=\arctan\left(\frac{1}{a}\right).$

Let

\begin{align*} P &= \arctan\left(\frac{1}{a+b}\right) \quad &&\Longrightarrow \quad& \tan P&=\frac{1}{a+b}\\ Q &= \arctan\left(\frac{1}{a+c}\right) \quad &&\Longrightarrow \quad& \tan Q&=\frac{1}{a+c} \\ R &= \arctan\left(\frac{1}{a}\right) \quad &&\Longrightarrow \quad& \tan R&=\frac{1}{a} \end{align*}

Note that since $a, b$ and $c$ are all positive numbers, $P$, $Q$ and $R$ all lie between $0$ and $\dfrac{\pi}{2}$.

Then we wish to show that $P+Q=R$, which is equivalent to showing that $\tan(P+Q)=\tan R$ (since $0 < P+Q < \pi$).

Using the compound angle formula, we also know that $\tan(P+Q)=\frac{\tan P + \tan Q}{1-\tan P\tan Q}.$ So we find that \begin{align*} \tan(P+Q) &= \dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{1-\dfrac{1}{(a+b)(a+c)}} \\ &= \dfrac{\Bigl(\dfrac{2a+b+c}{(a+b)(a+c)}\Bigr)}{\Bigl(\dfrac{a^2+ab+ac+bc-1}{(a+b)(a+c)}\Bigr)} \\ &= \dfrac{2a+b+c}{a^2+ab+ac+bc-1} \\ &= \dfrac{2a+b+c}{a^2+ab+ac+(a^2+1)-1}\quad\text{as bc=a^2+1}\\ &= \dfrac{2a+b+c}{2a^2+ab+ac}\\ &= \dfrac{2a+b+c}{a(2a+b+c)}\\ &= \dfrac{1}{a}\\ &= \tan R, \end{align*}

so we are done.

The positive numbers $p$, $q$, $r$, $s$, $t$, $u$ and $v$ satisfy $st=(p+q)^2+1, \qquad uv=(p+r)^2+1, \qquad qr=p^2+1.$

Prove that

\begin{align*} \arctan\left(\frac{1}{p+q+s}\right)&+\arctan\left(\frac{1}{p+q+t}\right)+\arctan\left(\frac{1}{p+r+u}\right)\\ &+\arctan\left(\frac{1}{p+r+v}\right)=\arctan\left(\frac{1}{p}\right). \end{align*}

Since $qr=p^2+1$, we can say immediately from the first part that

$\arctan\left(\frac{1}{p+q}\right)+\arctan\left(\frac{1}{p+r}\right)=\arctan\left(\frac{1}{p}\right).$

Now using the first part on the first term on the LHS (since $st=(p+q)^2+1$), and on the second term also (since $uv=(p+r)^2+1$), we have

$\arctan\left(\frac{1}{p+q+s}\right)+\arctan\left(\frac{1}{p+q+t}\right)+\arctan\left(\frac{1}{p+r+u}\right) +\arctan\left(\frac{1}{p+r+v}\right)=\arctan\left(\frac{1}{p}\right).$

Hence show that $\arctan\left(\frac{1}{13}\right)+\arctan\left(\frac{1}{21}\right)+\arctan\left(\frac{1}{82}\right)+\arctan\left(\frac{1}{187}\right)=\arctan\left(\frac{1}{7}\right).$

We are looking for values of $p$, $q$, $r$, $s$, $t$, $u$ and $v$ so that the above result takes this form.

We’ll begin by taking $p=7$ (to make the right hand side work) and then look for solutions consistent with this. We have the equations \begin{align*} 7+q+s &= 13 \qquad& \Longrightarrow&\quad& q+s &= 6\\ 7+q+t &= 21 \qquad& \Longrightarrow&\quad& q+t &= 14\\ 7+r+u &= 82 \qquad& \Longrightarrow&\quad& r+u &= 75\\ 7+r+v &= 187 \qquad& \Longrightarrow&\quad& r+v &= 180 \end{align*} and \begin{align*} st &= (7+q)^2+1\\ uv &= (7+r)^2+1\\ qr &= 7^2+1=50. \end{align*} Using the first two of these we find $s=6-q$ and $t=14-q$, and we can then substitute this into $st=(7+q)^2+1$ to eliminate $s$ and $t$, deducing \begin{align*} &&(6-q)(14-q)&=(q+7)^2+1&&\quad\\ &\Longrightarrow \quad& 84-20q+q^2&=q^2+14q+50\\ &\Longrightarrow \quad& 34q&=34\\ &\Longrightarrow \quad& q&=1. \end{align*}

It then follows that $s=5$, $t=13$ and $r=50$ since $qr=50$. Using this value for $r$ we also find that $u=25$ and $v=130$.

We must also double-check that all seven equations are consistent by subsituting into the equation we didn’t use: $uv = (p+r)^2+1$. This equation is satisfied, as $uv=25\times130 = 3250\quad\text{and}\quad (p+r)^2+1 = 57^2+1 = 3249+1=3250.$ So \begin{align*} p&=7,& q&=1,& r&=50,& s&=5,\\ t&=13,& u&=25,& v&=130 \end{align*}

gives us $\arctan\left(\frac{1}{13}\right)+\arctan\left(\frac{1}{21}\right)+\arctan\left(\frac{1}{82}\right)+\arctan\left(\frac{1}{187}\right)=\arctan\left(\frac{1}{7}\right).$