If $\tan (t)+\cot (t)=1$, then one of the values of the expression $\frac{1}{\sin (t)+\cos (t)}$ is _______.

Given the trigonometric equation:
$$\tan (t)+\cot (t)=1$$

Converting into sine and cosine functions:

$$\frac{\sin (t)}{\cos (t)}+\frac{\cos (t)}{\sin (t)}=1$$

$$\frac{{\sin }^2(t)+{\cos }^2(t)}{\sin (t)\cos (t)}=1$$

Since ${\sin }^2(t)+{\cos }^2(t)=1$:

$$\frac{1}{\sin (t)\cos (t)}=1⟹\sin (t)\cos (t)=1$$

Now let's find the value of the required expression. Let:

$$V=\frac{1}{\sin (t)+\cos (t)}$$

Squaring both sides:

$$V^2=\frac{1}{(\sin (t)+\cos (t){)}^2}$$

$$V^2=\frac{1}{{\sin }^2(t)+{\cos }^2(t)+2\sin (t)\cos (t)}$$

Substitute ${\sin }^2(t)+{\cos }^2(t)=1$ and $\sin (t)\cos (t)=1$:

$$V^2=\frac{1}{1+2(1)}=\frac{1}{3}$$

Taking the square root:

$$V=\pm \frac{1}{\sqrt{3}}$$

Thus, one possible value of the expression is $\frac{1}{\sqrt{3}}$.