If α and β are the solution of the equation, a tanθ + b secθ =

1.	If α and β are the solution of the equation, a tanθ + b secθ = c, then show that (α + β) = \(\frac{2ac}{a^2-c^2}\)
Answer» Given, α tanθ + b secθ = c ...(i) ⇒ (α tanθ − c)² = b² (1 + tan²θ) ⇒ (a² tan²b² ) − 2ac tanθ + c² = b² + b² tan²θ ⇒ (a² − b²) tan²θ − 2ac tanθ + c² − b² = 0 Since, α and β are roots of the equation (i), we have tan α + tan β = \(\frac{2ac}{a^2-b^2}\) and tan α.tan β = \(\frac{c^2-b^2}{a^2-b^2}\) Therefore, tan (α + β) = \(\frac{tan\alpha tan\beta}{1-tan\alpha tan\beta}\) =\(\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\) =\(\frac{2ac}{a^2-c^2}\)

If α and β are the solution of the equation, a tanθ + b secθ = c, then show that (α + β) = \(\frac{2ac}{a^2-c^2}\)

Answer»

Given, α tanθ + b secθ = c ...(i)

⇒ (α tanθ − c)² = b² (1 + tan²θ)

⇒ (a² tan²b² ) − 2ac tanθ + c² = b² + b² tan²θ

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

Since, α and β are roots of the equation (i),

we have

tan α + tan β = \(\frac{2ac}{a^2-b^2}\) and tan α.tan β = \(\frac{c^2-b^2}{a^2-b^2}\)

Therefore, tan (α + β) = \(\frac{tan\alpha tan\beta}{1-tan\alpha tan\beta}\)

=\(\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\)

=\(\frac{2ac}{a^2-c^2}\)