Given: a sec θ + b tan θ = x …….(1)

a tan θ + b sec θ = y …….(2)

Square equation (1) and (2) on both sides:

a² sec² θ + b² tan² θ + 2ab sec θ tan θ = x² …….(3)

a² tan² θ + b² sec² θ + 2ab sec θ tan θ = y² ……..(4)

Subtract equation (4) from (3):

[a² sec² θ + b² tan² θ + 2ab sec θ tan θ] – [a² tan² θ + b² sec² θ + 2ab sec θ tan θ] = x² – y²

⇒ a² (sec² θ – tan² θ) + b² (tan² θ – sec² θ) = x² – y²

⇒ a² – b² = x² – y² (∵sec² θ = 1 + tan² θ)

Hence, proved.