If tanA + sinA = m and tanA – sinA = n, then prove that (m² – n²)² = 16 nm

Given: -

(i) m = tanA + sinA

(ii) n = tanA – sinA

Proof:

Taking L.H.S

(m² – n²)² = (tanA + sinA)² – (tanA – sinA)²

⇒ (m² – n²)² = (tan²A + sin²A + 2tanAsinA – (tan²A + sin²A – 2tanAsinA))²

⇒ (m² – n²)² = (4tanAsinA)²

⇒ (m² – n²)² = 16tan²Asin²A

⇒ (m² – n²)² = 16tan²A(1 – cos²A)

⇒ (m² – n²)² = 16(tan²A – tan²Acos²A)

⇒ (m² – n²)² = 16(tan²A + sin²A) (tan²A – sin²A)

⇒ (m² – n²)² = 16 mn

Hence, Proved.