Question

If α and β are two solutions of the equation a tan x + b sec x = c, then find the values of sin (α + β) and cos (α + β).

Answer 1

$$\begin{gathered} a\tan x+b\sec x=c \\ \Rightarrow \left(c-a\tan x\right)=b\sec x \\ \Rightarrow {\left(c-a\tan x\right)}^2={\left(b\sec x\right)}^2 \\ \Rightarrow c^2+a^2{\tan }^{2}x-2ac\tan x=b^2{\sec }^{2}x \\ \Rightarrow c^2+a^2{\tan }^{2}x-2ac\tan x=b^2\left(1+{\tan }^{2}x\right) \\ \Rightarrow \left(a^2-b^2\right){\tan }^{2}x-2ac\tan x+\left(c^2-b^2\right)=0 \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{quadratic}\mathrm{in}\tan x. \\ \mathrm{It}\mathrm{has}\mathrm{two}\mathrm{solutions}\tan \alpha \mathrm{and}\tan \beta . \\ \tan \alpha +\tan \beta =\frac{2ac}{a^2-b^2} \\ \tan \alpha \times \tan \beta =\frac{c^2-b^2}{a^2-b^2} \\ \mathrm{Therefore},\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ =\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}} \\ =\frac{2ac}{a^2-c^2} \\ \mathrm{Hence},\sin \left(\alpha +\beta \right)=\frac{2ac}{a^2+c^2}\mathrm{and}\cos \left(\alpha +\beta \right)=\frac{a^2-c^2}{a^2+c^2}. \end{gathered}$$