From the top of a tower $h \space meter$ high, the angle of depression of two objects, which are in line to the foot of the tower is $\alpha$ and $\beta$ $( \beta > \alpha)$. Find the distance between the two objects.

$(\cot \alpha - \cot \beta) h \space meters$

Step by Step Explanation:
1. Let $AB$ be the tower of height $h \space meter$ and $x \space meter$ be the distance between the two objects $C$ and $D$.
As $\beta > \alpha$, $\beta$ will be the angle of depression of the point $D$ and $\alpha$ will be the angle of depression of the point $C.$

The situation given in the question is represented by the image given below.

2. In the right-angled triangle $ABD$, we have \begin{aligned} & \tan \beta = \dfrac { AB } { AD } \\ \implies & \tan \beta = \dfrac { h } { AD } \\ \implies & AD = \dfrac { h } { \tan \beta } \\ \implies & AD = h \space cot \beta && \bigg[\text{As,} \cot \beta = \dfrac { 1 } { \tan \beta } \bigg] && \ldots \text{(i)} \\ \end{aligned}
3. In right-angled triangle $ABC$, we have \begin{aligned} & \tan \alpha = \dfrac { AB } { AC } \\ \implies & \tan \alpha = \dfrac { h } { AC } \\ \implies & AC = \dfrac { h } { \tan \alpha } \\ \implies & AD + x = h \space \cot \alpha && \bigg[ \text{ As,} \cot \alpha = \dfrac { 1 } { \tan \alpha } \text{ and AC = AD + x } \bigg] && \ldots \text{(ii)} \\ \end{aligned}
4. Now, let us subtract $eq \space \text{(i)}$ from $eq \space \text{(ii)}$. \begin{aligned} & (AD + x) - AD = h \space cot \alpha - h \space cot \beta \\ \implies & x = (cot \alpha - cot \beta) h \\ \end{aligned}
5. Therefore, the distance between two objects is $(\cot \alpha - \cot \beta) h \space meters$.