Question

if u=log(tanx+tany+tanz) prove that sin2xdy/dx+sin2ydx/dy+sin2zdy/dx=2

Answer 1

Proof:

Given, u = log(tanx + tany + tanz)

⇒ eᵘ = tanx + tany + tanz ..... (1)

Taking partial derivatives with respect to x, y, z respectively we get

∂/∂x (eᵘ) = ∂/∂x (tanx + tany + tanz)

⇒ eᵘ ∂u/∂x = sec²x ..... (2)

∂/∂y (eᵘ) = ∂/∂y (tanx + tany + tanz)

⇒ eᵘ ∂u/∂y = sec²y ..... (3)

∂/∂z (eᵘ) = ∂/∂y (tanx + tany + tanz)

⇒ eᵘ ∂u/∂z = sec²z ..... (4)

We multiply (2), (3), (4) by sin2x, sin2y, sin2z respectively and add them. Then

sin2x (eᵘ ∂u/dx) + sin2y (eᵘ ∂u/∂y) + sin2z (eᵘ ∂u/∂z)

= sin2x sec²x + sin2y sec²y + sin2z sec²z

= (2 sinx cosx sec²x) + (2 siny cosy sec²y) + (2 sinz cosz sec²z)

= 2 (tanx + tany + tanz)

⇒ eᵘ (sin2x ∂u/∂x + sin2y ∂u/∂y + sin2z ∂u/∂z) = 2 eᵘ , by (1)

⇒ sin2x ∂u/∂x + sin2y ∂u/∂y + sin2z ∂u/∂z = 2

Hence proved.