## Task B-3.3.

If $\sin x + \cos x = a, |a| \leqslant \sqrt{2}$, what is $\frac{1 + \cos 2x}{\cot \frac{x}{2} - \tan \frac{x}{2}}$?
## Solution.

Let's write the expression $\frac{1+\cos 2 x}{\operatorname{ctg} \frac{x}{2}-\operatorname{tg} \frac{x}{2}}$ in a simpler form:

$$
\begin{aligned}
& \frac{1+\cos 2 x}{\operatorname{ctg} \frac{x}{2}-\operatorname{tg} \frac{x}{2}}=\frac{1+\cos 2 x}{\frac{1+\cos x}{\sin x}-\frac{1-\cos x}{\sin x}}= \\
& =\frac{\left(1+2 \cos ^{2} x-1\right) \sin x}{1+\cos x-1+\cos x}= \\
& =\frac{2 \cos ^{2} x \sin x}{2 \cos x}=\cos x \sin x
\end{aligned}
$$

It follows that

$$
\begin{array}{ll}
\sin x+\cos x=a \Rightarrow(\sin x+\cos x)^{2}=a^{2} & 1 \text { point } \\
\Rightarrow 1+2 \sin x \cos x=a^{2} & 1 \text { point } \\
\Rightarrow \sin x \cos x=\frac{a^{2}-1}{2} . & 1 \text { point }
\end{array}
$$