Aclock tower stands at the crossing of two roads whichpoint in the north-south and the east-westdirections. P, Q, R and S are points on the roads due north, east, south and west respectively, where the angles of elevation of the top of the tower are respectively, $$\alpha, \beta, \gamma$$ and $$\delta$$, Then $$\left(\frac{PQ}{RS}\right)^2$$ is equal to:

From the give question, we draw the diagram is given below

Now by the 3D view 

Let the height of tower = h

$$ \frac{h}{x} = \tan \alpha $$ ......(1)  ( alpha angle of elevation of tower with point P)

Similarly,  $$\frac{h}{y} = \tan \beta $$ ..... (2)

$$\frac {h}{z} = \tan \gamma $$ .......(3)

$$\frac{h}{a} = \tan \rho$$ -- (4)

From Equestion  (1)(2)(3)(4) 

$$x = \frac{h} {\tan \alpha} , a= \frac{h} {\tan \rho} , y= \frac{h} {\tan \beta} , z= \frac{h}{\tan \gamma} $$

Now PQ = $$ \sqrt {x^2 +y^2} $$

           = $$\sqrt{ \frac {h^2}{\tan^2 \alpha} + \frac{h^2}{ \tan^2 \beta} }$$  ...... Equestion (5) 

RS = $$\sqrt { z^2 + a^2} $$ 

   = $$\sqrt{\frac{h^2} {\tan^2 \gamma} + \frac{h^2} {\tan^2 \rho } }$$ ......... equestion (6)

Equestion (5) divided by Equestion (6)

$$ (\frac{PQ}{RS} )^2= \frac{ \cot^2 \alpha + \cot^2 \beta} { \cot^2 \gamma + \cot^2 \rho } $$

therefore Option (C) Ans