From the top of a building $h \ m$ high,the angle of elevation of the top of a pole is found to be $\alpha$,while the angle of depression of the base of the pole is found to be $\beta$. Prove that the height of the pole is $h(1 + \tan \alpha \cdot \cot \beta) \ m$.

(N/A) Let the building be $AB$ with height $h$ and the pole be $CD$ with height $H$. Let the distance between the building and the pole be $x$.
In $\triangle ABD$,$\angle ADB = \beta$ (angle of depression). Thus,$\tan \beta = \frac{AB}{BD} = \frac{h}{x}$,which implies $x = h \cot \beta$.
Now,consider the triangle formed by the top of the building and the top of the pole. Let $E$ be a point on $CD$ such that $AE \perp CD$. Then $AE = BD = x$ and $ED = AB = h$.
In $\triangle AEC$,$\angle EAC = \alpha$ (angle of elevation). Thus,$\tan \alpha = \frac{EC}{AE} = \frac{EC}{x}$,which implies $EC = x \tan \alpha$.
Substituting $x = h \cot \beta$,we get $EC = (h \cot \beta) \tan \alpha = h \tan \alpha \cot \beta$.
The total height of the pole $H = ED + EC = h + h \tan \alpha \cot \beta = h(1 + \tan \alpha \cot \beta)$.