If $$\delta $$ is the declination of the Polaris and $$\lambda $$ is the latitude of the place, the azimuth of the Polaris, is A. $$\frac{{\cos \delta }}{{\cos \lambda }}$$ B. $$\frac{{\cos \left( {{{90}^ \circ } – \delta } \right)}}{{\cos \left( {{{90}^ \circ } – \lambda } \right)}}$$ C. $$\frac{{\sin \left( {{{90}^ \circ } – \delta } \right)}}{{\sin \left( {{{90}^ \circ } – \lambda } \right)}}$$ D. $$\frac{{\tan \left( {{{90}^ \circ } + \delta } \right)}}{{\tan \left( {{{90}^ \circ } + \lambda } \right)}}$$

The correct answer is $\boxed{\frac{{\cos \left( {{{90}^ \circ } – \delta } \right)}}{{\cos \left( {{{90}^ \circ } – \lambda } \right)}}}$.

The azimuth of a celestial object is the angle between the north celestial pole and the object, measured eastward from north along the horizon. The declination of a celestial object is its angular distance from the celestial equator, measured north or south.

The azimuth of Polaris is the angle between the north celestial pole and Polaris, measured eastward from north along the horizon. The declination of Polaris is $+89.27^\circ$. The latitude of a place is the angle between the horizon and the Earth’s axis, measured north or south.

The azimuth of Polaris can be calculated using the following formula:

$$\text{Azimuth of Polaris} = \arccos \left( \frac{\cos \delta}{\cos \lambda} \right)$$

where $\delta$ is the declination of Polaris and $\lambda$ is the latitude of the place.

The other options are incorrect because they do not take into account the declination of Polaris.