A person observes the angle of elevation of the peak of a hill from a station to be α.

Suppose, AB is a peak whose height above the ground is t+x.

In $\triangle D F C$,

$\sin \beta=\frac{x}{c}$

$\Rightarrow x=c \sin \beta$

and,

$\tan \beta=\frac{x}{y}$

$\Rightarrow \mathrm{y}=\frac{x}{\tan \beta}=\frac{c \sin \beta}{\sin \beta} \times \cos \beta=c \cos \beta \quad \ldots$ (1)

In $\Delta A D E$,

$\tan \gamma=\frac{t}{z}$

$\Rightarrow z=t \cot \gamma \quad \ldots(2)$

In $\Delta A B C$,

$\tan \alpha=\frac{t+x}{y+z}$

$\Rightarrow t+x=(c \cos \beta \tan \alpha+t \cot \gamma) \tan \alpha \quad$ (from (1) and (2) $)$

$\Rightarrow t-t \cot \gamma \tan \alpha=c \cos \beta \tan \alpha-c \sin \beta \quad(\because x=c \sin \beta)$

$\Rightarrow t\left(1-\frac{\sin \alpha \cos \gamma}{\cos \alpha \sin \gamma}\right)=c\left(\frac{\cos \beta \sin \alpha-\cos \alpha \sin \beta}{\cos \alpha}\right)$

$\Rightarrow t\left(\frac{\sin \gamma \cos \alpha-\sin \alpha \cos \gamma}{\cos \alpha \sin \gamma}\right)=c \frac{\sin (\alpha-\beta)}{\cos \alpha}$

$\Rightarrow t \frac{\sin (\gamma-\beta)}{\cos \alpha \sin \gamma}=c \frac{\sin (\alpha-\beta)}{\cos \alpha}$

$\Rightarrow t=c \frac{\sin \gamma \sin (\alpha-\beta)}{\sin (\gamma-\beta)} \quad \ldots(3)$

Now,

$A B=t+x=c \frac{\sin \gamma \sin (\alpha-\beta)}{\sin (\gamma-\beta)}+c \sin \beta \quad($ using $(3))$

$=c\left(\frac{\sin \gamma \sin (\alpha-\beta)}{\sin (\gamma-\beta)}+\sin \beta\right)$

$=c\left[\frac{\sin \gamma \sin (\alpha-\beta)+\sin \beta \sin (\gamma-\beta)}{\sin (\gamma-\beta)}\right]$

$=c\left[\frac{\sin \gamma \sin \alpha \cos \beta-\sin \beta \sin \gamma \cos \alpha+\sin \beta \sin \gamma \cos \alpha-\sin \beta \cos \gamma \sin \alpha}{\sin (\gamma-\beta)}\right]$

$=c\left[\frac{\sin \gamma \sin \alpha \cos \beta-\sin \beta \cos \gamma \sin \alpha}{\sin (\gamma-\beta)}\right]$

$=c\left[\frac{\sin \alpha \sin (\gamma-\beta)}{\sin (\gamma-\beta)}\right]$

$=\frac{c \sin \alpha \sin (\gamma-\beta)}{\sin (\gamma-\beta)}$

Hence proved.