## Task B-3.5.

If $\operatorname{tg} x+\operatorname{ctg} x=-2, x \neq k \frac{\pi}{2}, k \in \mathbb{Z}$, what is $(\sin x-\cos x)^{10}$ ?
## First solution.

From the equality $\operatorname{tg} x+\operatorname{ctg} x=-2$ it follows that

$$
\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=-2
$$

or

$$
\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}=-2
$$

Since $\sin ^{2} x+\cos ^{2} x=1$, it follows that

$$
\cos x \sin x=-\frac{1}{2}
$$

Now we have:

$$
\begin{array}{cc}
(\sin x-\cos x)^{10}=\left((\sin x-\cos x)^{2}\right)^{5}= & 1 \text { point } \\
\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)^{5}=(1-2 \sin x \cos x)^{5}= & 1 \text { point } \\
\left(1-2 \cdot\left(-\frac{1}{2}\right)\right)^{5}=32 & 1 \text { point }
\end{array}
$$