Simplify the following expression:

$$
\frac{\frac{a+b}{1-a b}+\frac{c-a}{1+c a}}{1-\frac{a+b}{1-a b} \cdot \frac{c-a}{1+c a}}
$$

Explain the result using trigonometric identities.
Let's expand the fraction by $(1-a b)(1+c a)$. (This product is non-zero, because if any of its factors were to vanish, the expression would lose its meaning.) Thus, after the anticipated simplifications, the remaining 4-4 terms can be transformed into the product of two binomials by simple factoring. The common factor $1+a^{2}$ in the numerator and denominator is certainly positive, and can thus be simplified, so the expression is:

$$
K=\frac{b+a^{2} c+c+a^{2} b}{1-a^{2} b c-b c+a^{2}}=\frac{(b+c)\left(1+a^{2}\right)}{(1-b c)\left(1+a^{2}\right)}=\frac{b+c}{1-b c}
$$

From this, we see that $K$ would not be defined if $1-b c=0$.

The quotients

$$
\frac{a+b}{1-a b}=d \quad \text { and } \quad \frac{c-a}{1+c a}=e
$$

occurring twice in the expression, as well as the form of $K$ when these are substituted,

$$
K=\frac{d+e}{1-d e}
$$

remind us of the addition formula for the tangent function,

$$
\operatorname{tg}(x+y)=\frac{\operatorname{tg} x+\operatorname{tg} y}{1-\operatorname{tg} x \operatorname{tg} y}
$$

from which the known identity for $\operatorname{tg}(x-y)$ follows by considering $\operatorname{tg}(-y)=-\operatorname{tg} y$.

If we find the angles $\alpha, \beta, \gamma$ between $-90^{\circ}$ and $90^{\circ}$ such that

$$
\operatorname{tg} \alpha=a, \quad \operatorname{tg} \beta=b, \quad \operatorname{tg} \gamma=c
$$

then, from (1) and (3),

$$
d=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\operatorname{tg}(\alpha+\beta) \quad \text { and } \quad \mathrm{e}=\operatorname{tg}(\gamma-\alpha)
$$

Substituting these into (2), similarly,

$$
K=\frac{\operatorname{tg}(\alpha+\beta)+\operatorname{tg}(\gamma-\alpha)}{1-\operatorname{tg}(\alpha+\beta) \operatorname{tg}(\gamma-\alpha)}=\operatorname{tg}[(\alpha+\beta)+(\gamma-\alpha)]=\operatorname{tg}(\beta+\gamma)
$$

From this, by transforming (3) and finally (4), we can return to the original notation without intermediate calculations and arrive at the final result of the original transformation.

We must note that the angles $\alpha, \beta, \gamma$ exist for any $a, b, c$, because the function $y=\operatorname{tg} x$ takes all values. However, if either

$$
1-a b=0
$$

or

$$
1+c a=0
$$

were to hold, or both of these equations were to hold, then the trigonometric transformation would not be valid. Indeed, from (7), $a b=1$, so neither $a$ nor $b$ is $0$, $b=1 / a$, and thus $\operatorname{tg} \beta=\operatorname{ctg} \alpha$, meaning $\alpha$ and $\beta$ are complementary angles, their sum is $90^{\circ}$, or $\alpha+\beta=-90^{\circ}$, and their tangent is undefined; similarly, if (8) holds, $c=-1 / a, \operatorname{tg} \gamma=-\operatorname{ctg} \alpha=\operatorname{ctg}(-\alpha)$, so $\gamma+(-\alpha)= \pm 90^{\circ}$, and (5) is undefined. In these cases, the trigonometric transformation cannot even begin, just as the algebraic one cannot. And the transformation started cannot be completed with (6) if we find that $\beta+\gamma=90^{\circ}$, in accordance with the exclusionary remark for the case $1-b c=0$.

Patrik Fazekas (Mosonmagyaróvár, Kossuth L. Gymnasium II. grade)