Question

If a sec θ + b tan θ = m and b sec θ + a tan θ = n, prove that a² + n² = b² + m²

Answer 1

a secθ + b tanθ = m     ....(i)

b secθ + a tanθ = n     ....(ii)

Squaring both sides of the equations

a² sec²θ + b² tan²θ + 2ab secθ tanθ = m²
b² sec²θ + a² tan²θ + 2ab secθ tanθ = n²
−            −                −                        −         
a² (sec²θ − tan²θ) − b² (sec²θ − tan²θ) = m² − n²

⇒ a² × 1 − b² × 1 = m² − n²

⇒ a² + n² = b² + m²