### From a window $h \space meters$ high above the ground in a street, the angles of elevation and depression of the top and foot of the other house on the opposite side of the street are $\alpha$ and $\beta$ respectively. Show that the height of the opposite house is $h (1 + \tan \alpha \space cot \beta) \space meters$.

Step by Step Explanation:
1. Let $W$ be the window and $AB$ be the house on the opposite side with height $(h + h') \space meters$.

The figure below shows the given situation.
2. In the right-angled triangle $AWP$, we have \begin{aligned} & \cot \alpha = \dfrac { WP } { AP } \\ \implies & \cot \alpha = \dfrac { WP } { h' } \\ \implies & WP = h' \cot \alpha && \ldots \text{(i)} \end{aligned}
3. In right-angled triangle $WPB$, we have \begin{aligned} & \cot \beta = \dfrac { WP } { BP } \\ \implies & \cot \beta = \dfrac { WP } { h } \\ \implies & WP = h \cot \beta && \ldots \text{(ii)} \end{aligned}
4. On comparing $eq \space \text{(i)}$ and $eq \space \text{(ii)}$, we get \begin{aligned} & h' \cot \alpha = h \cot \beta \\ \implies & h' = \dfrac { h \cot \beta } { \cot \alpha } = h \tan \alpha \cot \beta \end{aligned}
5. Thus, height of the house = $h + h' = h + h \tan \alpha \cot \beta = h (1 + \tan \alpha \cot \beta) \space meters$.