The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be α. After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to β. Then the height (in metres) of the tower is :

Let h be the height of the tower and L be the distance of point P from the foot of the tower.
Then, in the first case, we have tanα = h/L
In the second case, after moving 2 meters towards the foot of the tower, the distance from the foot of the tower becomes L-2. Then we have tanβ = h/(L-2)
From the first equation, L = h/tanα = h cotα
Substituting this into the second equation, we get:
tanβ = h/(h cotα - 2)
Rearranging the equation, we get:
h cotα - 2 = h/tanβ
h cotα - h/tanβ = 2
h(cotα - 1/tanβ) = 2
h(cosα/sinα - sinβ/cosβ) = 2
h(cosαcosβ - sinαsinβ)/(sinαcosβ) = 2
h cos(α+β)/(sinαcosβ) = 2
Also, tan(β-α) = (tanβ - tanα)/(1 + tanαtanβ) = (h/(L-2) - h/L)/(1 + h²/L(L-2))
We have tanα = h/L and tanβ = h/(L-2)
Then tanβ - tanα = h/(L-2) - h/L = hL - h(L-2)/L(L-2) = 2h/L(L-2)
1 + tanαtanβ = 1 + (h²/L(L-2))
Therefore, tan(β-α) = 2h/L(L-2) / (1 + h²/L(L-2)) = 2h/(L(L-2) + h²)
From tanα = h/L, L = h/tanα, and from tanβ = h/(L-2), L-2 = h/tanβ, then L = h/tanβ + 2
Equating the two expressions for L:
h/tanα = h/tanβ + 2
h(1/tanβ - 1/tanα) = 2
h(tanα - tanβ)/(tanαtanβ) = 2
h(sinα/cosα - sinβ/cosβ) = 2
h(sinαcosβ - cosαsinβ)/(cosαcosβ) = 2
h sin(α-β)/(cosαcosβ) = 2
h = 2cosαcosβ/sin(α-β)
Solving for h:
h(cotα - cotβ) = 2
h = 2 / (cotα - cotβ) = 2 / ((cosα/sinα) - (cosβ/sinβ)) = 2sinαsinβ / (cosαsinβ - sinαcosβ) = 2sinαsinβ / sin(β-α)
Therefore, h = 2sinαsinβ/sin(β-α)