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

## Suggestion

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

Do we know what $\tan(A+B)$ is?

Could we use $P=\arctan\left(\dfrac{1}{a+b}\right) \implies \tan P=\dfrac{1}{a+b}$?

Can we simplify the LHS of the first equation with this?

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*}

Can we use the first part of the question? Could we use it twice?

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

Can we use simultaneous equations to find suitable values for the various letters?