2011-II-50

Ans: B
Consider $\Delta BFG$,

$\begin{array}{rcl}
\tan y & = & \dfrac{BG}{FG} \\
BG & = & FG\tan y
\end{array}$

Consider $\Delta FGH$,

$\begin{array}{rcl}
\tan x & = & \dfrac{FG}{GH} \\
GH & = & \dfrac{FG}{\tan x}
\end{array}$

Consider $\Delta BGH$,

$\begin{array}{rcl}
\tan z & = & \dfrac{GH}{BG} \\
& = & \dfrac{FG}{\tan x} \times \dfrac{1}{FG \tan y} \\
& = & \dfrac{1}{\tan x \tan y}
\end{array}$