A window of a house is h m above the ground.

Let the height of the other house $=O Q=H$

and $\quad O B=M W=x \mathrm{~m}$

Given that, height of the first house $=W B=h=M O$

and $\angle Q W M=\alpha, \angle O W M=\beta=\angle W O B$

[alternate angle]

Now, in $\Delta W O B, \quad \tan \beta=\frac{W B}{O B}=\frac{h}{x}$

$\Rightarrow$ $x=\frac{h}{\tan \beta}$ ...(i)

And in $\triangle Q W M$, $\tan \alpha=\frac{Q M}{W M}=\frac{O Q-M O}{W M}$

$\Rightarrow$ $\tan \alpha=\frac{H-h}{x}$

$\Rightarrow$ $x=\frac{H-h}{\tan \alpha}$ ... (ii)

From Eqs. (i) and (ii),

$\frac{h}{\tan \beta}=\frac{H-h}{\tan \alpha}$

$\Rightarrow \quad h \tan \alpha=(H-h) \tan \beta$

$\Rightarrow \quad h \tan \alpha=H \tan \beta-h \tan \beta$

$\Rightarrow \quad H \tan \beta=h(\tan \alpha+\tan \beta)$

$\therefore$ $H=h\left(\frac{\tan \alpha+\tan \beta}{\tan \beta}\right)$

$=h\left(1+\tan \alpha \cdot \frac{1}{\tan \beta}\right)=h(1+\tan \alpha \cdot \cot \beta)\left[\because \cot \theta=\frac{1}{\tan \theta}\right]$

Hence, the required height of the other house is $h(1+\tan \alpha \cdot \cot \beta)$

Hence proved.