## Task A-3.2.

If $(1+\sin t)(1+\cos t)=\frac{5}{4}$, find $\sin t+\cos t$.
## Solution.

Let $a=\sin t+\cos t$. Then $a^{2}=1+2 \sin t \cos t$.

By expanding the given equality, we get

$$
\begin{gathered}
1+(\sin t+\cos t)+\sin t \cos t=\frac{5}{4} \\
1+a+\frac{1}{2}\left(a^{2}-1\right)=\frac{5}{4}
\end{gathered}
$$

From here, by simplifying, we obtain the quadratic equation $2 a^{2}+4 a-3=0$

whose solutions are

$$
a=\frac{-4 \pm \sqrt{4^{2}+4 \cdot 3 \cdot 2}}{2 \cdot 2}=\frac{-2 \pm \sqrt{10}}{2}=-1 \pm \frac{\sqrt{10}}{2}
$$

Since $-1-\frac{\sqrt{10}}{2} \approx-2.58$ is less than -2, and $\sin t+\cos t$ takes values in the interval $[-\sqrt{2}, \sqrt{2}]$, this solution must be discarded.

Therefore, the only solution is $a=-1+\frac{\sqrt{10}}{2} \approx 0.58$.