Question

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

Answer 1

Given that atanθ + bsecθ = c or asinθ + b = c cos θ

Using the identities,

sin θ = `(2tan  theta/2)/(1 + tan^2  theta/2)` and cos θ = `(1 - tan^2  theta/2)/(1 + tan^2  theta/2)`

We have, `(a(2tan  theta/2))/(1 + tan^2  theta/2) + b = (c(1 - tan^2  theta/2))/(1 + tan^2  theta/2)`

or `(b + c) tan^2  theta/2 + 2a tan  theta/2 + b - c` = 0

The above equation is quadratic in `tan  theta/2` and hence `tan  alpha/2` and `tan  beta/2` are the roots of this equation.

Therefore, `tan  alpha/2 + tan  beta/2 = (-2a)/(b + c)` and `tan  alpha/2  tan  beta/2 - (b - c)/(b + c)` 

Using the identity `tan(alpha/2 + beta/2) = (tan  alpha/2 + tan  beta/2)/(1 - tan  alpha/2 tan   beta/2)`

We have, `tan(alpha/2 + beta/2) = ((-2a)/(b + c))/(1 - (b - c)/(b + c))`

= `(-2a)/(2c) = (-a)/c`  .....(1)

Again, using another identity

`tan 2 (alpha + beta)/2 = (2tan  (alpha + beta)/2)/(1 - tan^2  (alpha + beta)/2)`

We have tan (α + β) = `(2(- a/c))/(1 - a^2/c^2)`

= `(2ac)/(a^2 - c^2)`  ......[From (1)]

Alternatively, given that a tanθ + b secθ = c

⇒ (a tanθ – c)² = b² (1 + tan²θ) 

⇒ a² tan²θ – 2ac tanθ + c² = b² + b² tan²θ

⇒ (a² – b²) tan²θ – 2ac tanθ + c² – b² = 0  ......(1)

Since α and β are the roots of the equation (1)

So tanα + tanβ = `(2ac)/(a^2 - b^2)`

And tanα tanβ = `(c^2 - b^2)/(a^2 - b^2)`

Therefore,  tan (α + β) = `(tan  alpha + tan beta)/(1 - tan alpha tan beta)`

= `((2ac)/(a^2 - b^2))/((c^2 - b^2)/(a^2 - b^2))`

= `(2ac)/(a^2 - c^2)`