From the top of a tower,the angles of depression of points $A$ and $B$ in the same direction of the tower are found to be $\theta$ and $(90^\circ - \theta)$ respectively. If $B$ is nearer the tower than $A$ and $AB = a$,then prove that the height of the tower is $\frac{a \tan \theta}{1 - \tan^2 \theta}$.

(A) Let $h$ be the height of the tower $PQ$,where $P$ is the top and $Q$ is the base. Let $Q$ be the origin on the ground. Points $A$ and $B$ lie on the same line from $Q$. Let $QB = x$ and $QA = x + a$.
In $\triangle PQB$,$\tan(90^\circ - \theta) = \frac{h}{x} \implies \cot \theta = \frac{h}{x} \implies x = h \tan \theta$.
In $\triangle PQA$,$\tan \theta = \frac{h}{x + a} \implies x + a = \frac{h}{\tan \theta} = h \cot \theta$.
Substituting $x = h \tan \theta$ into the second equation: $h \tan \theta + a = h \cot \theta$.
$a = h(\cot \theta - \tan \theta) = h(\frac{1}{\tan \theta} - \tan \theta) = h(\frac{1 - \tan^2 \theta}{\tan \theta})$.
Therefore,$h = \frac{a \tan \theta}{1 - \tan^2 \theta}$.