$A$ window of a house is $h$ meters above the ground. From the window,the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$,respectively. Prove that the height of the other house is $h(1 + \tan \alpha \cot \beta)$ meters.

(N/A) Let the height of the other house be $OQ = H$ and the distance between the two houses be $OB = MW = x \text{ m}$.
Given that,the height of the first house is $WB = h = MO$.
The angle of elevation of the top is $\angle QWM = \alpha$ and the angle of depression of the bottom is $\angle OWM = \beta$.
Since $WM$ is parallel to $BO$,$\angle WOB = \angle OWM = \beta$ (alternate interior angles).
In $\triangle WOB$,$\tan \beta = \frac{WB}{OB} = \frac{h}{x} \implies x = \frac{h}{\tan \beta} \dots (i)$.
In $\triangle QWM$,$\tan \alpha = \frac{QM}{WM} = \frac{OQ - MO}{WM} = \frac{H - h}{x} \implies x = \frac{H - h}{\tan \alpha} \dots (ii)$.
Equating $(i)$ and $(ii)$:
$\frac{h}{\tan \beta} = \frac{H - h}{\tan \alpha}$
$h \tan \alpha = (H - h) \tan \beta$
$h \tan \alpha = H \tan \beta - h \tan \beta$
$H \tan \beta = h \tan \alpha + h \tan \beta$
$H \tan \beta = h(\tan \alpha + \tan \beta)$
$H = h \left( \frac{\tan \alpha + \tan \beta}{\tan \beta} \right)$
$H = h \left( \frac{\tan \alpha}{\tan \beta} + 1 \right)$
$H = h(1 + \tan \alpha \cot \beta)$.
Hence,the height of the other house is $h(1 + \tan \alpha \cot \beta)$ meters.