If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan(α + β) = (2ac) / (a² − c²).

If α and β are the solutions of the equation atan θ + bsec θ = c

Given that atanθ + bsecθ = c

(a tan θ − c)² = b²(1 + tan²θ)

⇒ a²tan²θ − 2ac tan θ + c² = b² + b²tan²θ

⇒ (a² − b²)tan²θ − 2ac tan θ + (c² − b²) = 0

Since, tan α and tan β are roots of the equation, so

tan α + tan β = (2ac) / (a² − b²) and tan α × tan β = (c² − b²) / (a² − b²)

Therefore, tan(α + β) = (tan α + tan β) / (1 − tan α × tan β)

= ((2ac) / (a² − b²)) / (1 − ((c² − b²) / (a² − b²)))

= (2ac) / (a² − c²)