Question should be:

If the apparent dips at two planes, perpendicular to each other, are δ ₁ and δ ₂ , then if δ is the real dip, we need to prove  cot ² δ = cot ² δ ₁ + cot ² δ ₂

_Answer:

Let B_H and B_V be the horizontal and vertical components of earths magnetic field vector B.

Then tan δ = B_V / B_H  ----------------Eqn (1)

Plane 1:

Let us consider the horizontal component B_H alone. The pictorial representation indicates that the apparent dip δ _1,  in the plane 1 makes an angle Ɵ with the magnetic meridian.

The B_V component remains as it is. But the B_H component from the figure shows, it is B_H cos Ɵ

Hence, Eqn (1) becomes tan δ₁ = B_V / B_H  cos Ɵ  -------Eqn (2)

Substitute in Eqn (2) that B_V = B_H tan δ

Eqn (2) becomes tan δ₁ = B_H tan δ / B_H cos Ɵ 

                                   = tan δ /  cos Ɵ 

cos Ɵ  = tan δ / tan δ₁ = tan δ cot δ₁ ---------------Eqn (3)

Plane 2:

The angle made by the plane 2 with the horizontal component will be 90 – Ɵ

tan δ₂ = B_V / B_H  cos(90- Ɵ)  = B_V / B_H  sin Ɵ  -------Eqn (4)

Substitute in Eqn (4) that B_V = B_H tan δ

Eqn (4) becomes tan δ₂ = B_H tan δ / B_H sin Ɵ 

                                   = tan δ /  sin Ɵ 

sin Ɵ  = tan δ / tan δ₂ = tan δ cot δ₂ ---------------Eqn (5)

Square and add equations (3) and (4)

cos² Ɵ  + sin ² Ɵ = tan² δ cot² δ_{1 +} tan² δ cot² δ₁

• = tan² δ ( cot² δ₁ + cot² δ₂)
• = 1/ cot² δ * [ ( cot² δ₁ + cot² δ₂)]

cot² δ =  cot² δ₁ + cot² δ₂ which is proved.