If θ₁, θ₂, θ₃, ..., θ_n are in A.P., whose common difference is d, show that Sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + ... + sec θ_n–1 sec θ_n

we have to prove that

sec θ₁ sec θ₂ + sec θ₂ sec θ₃ + ... + sec θ_n–1 sec θ_n =

⇒ sind(secθ₁ secθ₂ + secθ₂ secθ₃ + ... + sec θ_n–1 sec θ_n) = tan θ_n – tan θ₁

Consider LHS

Now we have to find value of d in terms of θ so that further simplification can be made

As θ₁, θ₂, θ₃, ..., θ_n are in AP having common difference as d

Hence

θ₂ – θ₁ = d, θ₃ – θ₂ = d, …, θ_n – θ_n-1 = d

Take sin on both sides

sin(θ₂ – θ₁) = sind, sin(θ₃ – θ₂) = sind, …, sin(θ_n – θ_n-1) = sind

Substitute appropriate value of sind for each term in LHS

We know that sin(a – b) = sinacosb – cosasinb

= tan θ₂ – tan θ₁ + tan θ₃ – tan θ₂ + … + tan θ_n – tan θ_n-1

= - tan θ₁ + tan θ_n

= tan θ_n - tan θ₁

⇒ LHS = RHS

Hence proved