If α and β are solutions of a tanx + b secx = c, Find sin(α+β) and cos(α+β)

Question

If

\[ a\tan x+b\sec x=c \]

and \[ \alpha,\beta \] are two solutions of the equation, find:

\[ \sin(\alpha+\beta) \quad \text{and} \quad \cos(\alpha+\beta) \]

Solution

Given,

\[ a\tan x+b\sec x=c \]

\[ a\frac{\sin x}{\cos x} + b\frac{1}{\cos x} =c \]

\[ a\sin x+b=c\cos x \]

\[ a\sin x-c\cos x=-b \]

Let

\[ a=r\cos\theta, \qquad c=r\sin\theta \]

Then,

\[ r\sin(x-\theta)=-b \]

\[ \sin(x-\theta)=-\frac{b}{r} \]

Hence the two solutions are:

\[ \alpha-\theta=\sin^{-1}\left(-\frac{b}{r}\right) \]

and

\[ \beta-\theta=\pi-\sin^{-1}\left(-\frac{b}{r}\right) \]

Adding,

\[ \alpha+\beta-2\theta=\pi \]

\[ \alpha+\beta=\pi+2\theta \]

Therefore,

\[ \sin(\alpha+\beta) = \sin(\pi+2\theta) \]

\[ = -\sin2\theta \]

\[ = -2\sin\theta\cos\theta \]

\[ = -2\cdot\frac{c}{r}\cdot\frac{a}{r} \]

\[ = -\frac{2ac}{a^2+c^2} \]

Also,

\[ \cos(\alpha+\beta) = \cos(\pi+2\theta) \]

\[ = -\cos2\theta \]

\[ = -(\cos^2\theta-\sin^2\theta) \]

\[ = -\left(\frac{a^2}{r^2}-\frac{c^2}{r^2}\right) \]

\[ = \frac{c^2-a^2}{a^2+c^2} \]

Therefore,

\[ \boxed{ \sin(\alpha+\beta) = -\frac{2ac}{a^2+c^2} } \]

and

\[ \boxed{ \cos(\alpha+\beta) = \frac{c^2-a^2}{a^2+c^2} } \]