This is the basic and conceptual question from trigonometry, topic – trigonometric identities in which we have given that (tanθ+sinθ)=m and also (tanθ-sinθ)=n, and we need to prove that (m²-n²)² = 16mn
RS Aggarwal, Class 10, chapter 13B, question no 7
(tan θ + sin θ) = m
and (tan θ − sin θ) = n
To Prove: (m² − n²)² = 16mn
L.H.S. = (m² − n²)²
= [(tan θ + sin θ)² – (tan θ − sin θ)²]² = (4tan θ sin θ)²
= 16 tan²θ sin²θ…….(1)
R.H.S.
= 16mn
= 16(tan θ + sin θ)(tan θ − sin θ)
= 16(tan²θ − sin²θ)
= 16 [{sin²θ (1-cos²θ)/cos²θ]
= 16 × sin²θ/cos²θ × (1-cos²θ)
= 16 tan²θ sin²θ …(2)
From (1) and (2) L.H.S. = R.H.S.