Given: tan θ + sin θ = m …….(1)

tan θ – sin θ = n …….(2)

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

tan² θ + sin² θ + 2 sin θ tan θ = m² …….(3)

tan² θ + sin² θ – 2 sin θ tan θ = n² ……..(4)

Subtract equation (4) from (3):

[tan² θ + sin² θ + 2 sin θ tan θ] – [tan² θ + sin² θ – 2 sin θ tan θ] = m² – n²

⇒ 4sin θ tan θ = m² – n²

Square both sides:

⇒ 16 sin² θ tan² θ = (m² – n²)²

Therefore, (m² – n²)² = 16 sin² θ tan² θ

Also, 16mn = 16 × (tan θ + sin θ) × (tan θ – sin θ)

= 16 (tan² θ – sin² θ)

= 16[(sin² θ/ cos² θ) – sin² θ]

= 16[sin² θ ]

= 16 sin² θ(sin² θ/ cos² θ)

= 16 sin² θ tan² θ

Therefore, (m² – n²)² = 16mn

Hence, proved.