Evaluate the sum

$$
\frac{1}{1+\tan 1^{\circ}}+\frac{1}{1+\tan 2^{\circ}}+\frac{1}{1+\tan 3^{\circ}}+\cdots+\frac{1}{1+\tan 89^{\circ}}
$$

(The tangent $(\tan )$ of an angle $\alpha$ is the ratio $B C / A C$ in a right triangle $A B C$ with $\angle C=90^{\circ}$ and $\angle A=\alpha$, and its value does not depend on the triangle used.)
By examining a triangle with angles $x, 90-x$, and 90 , it is not hard to see the trigonometric identity

$$
\tan \left(90^{\circ}-x\right)=\frac{1}{\tan x} .
$$

Let us pair up the terms

$$
\frac{1}{1+\tan x} \text { and } \frac{1}{1+\tan \left(90^{\circ}-x\right)}
$$

for $x=1,2,3, \ldots, 44$. The sum of such a pair of terms is

$$
\begin{aligned}
\frac{1}{1+\tan x}+\frac{1}{1+\tan \left(90^{\circ}-x\right)} & =\frac{1}{1+\tan x}+\frac{1}{1+\frac{1}{\tan x}} \\
& =\frac{1}{1+\tan x}+\frac{\tan x}{\tan x+1} \\
& =\frac{1+\tan x}{1+\tan x}=1
\end{aligned}
$$

There is one more term in the sum, the middle term:

$$
\frac{1}{1+\tan 45^{\circ}}=\frac{1}{1+1}=\frac{1}{2} .
$$

Therefore the sum is $44(1)+1 / 2=89 / 2$.