From an aeroplane vertically above a straight horizontal plane, the angles of depression of two consecutive kilometres stones on the opposite sides of the aeroplane are found to be "alpha" and "beta". show that the height of the aeroplane is:
(tan alpha * tan beta)/(tan alpha + tan beta).

You have to build the triangles.

They are such that:
h is the common height
x is the horizontal distance from the plane to one stone
Beta is the angle between x and the  hypotenuse

Then in this triangle: tan(beta) = h / x (1)

1 - x is the horizontal distance from the plane to the other stone
alfa is the angle between 1 - x  and h

Then, in this triangle: tan (alfa) = h / [1 -x ] (2)

from (1) , x = h / tan(beta)

Substitute this value in (2)

tan(alfa) = h / { [ 1 - h / tan(beta)] } =>

{ [ 1 - h / tan(beta) ] } tan(alfa) = h

[tan(beta) - h] tan(alfa) = h*tan(beta)

tan(beta)tan(alfa) - htan(alfa) = htan(beta)

h [tan(alfa) + tan(beta) ] = tan(beta) tan (alfa)

h = tan(beta)*tan(alfa) / (t an(alfa)  + tan(beta) )