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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)`.
Solution
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)`
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