If α and β are two solutions of the equation a tan x + b sec x = c,

$a \tan x+b \sec x=c$

$\Rightarrow(c-a \tan x)=b \sec x$

$\Rightarrow(c-a \tan x)^{2}=(b \sec x)^{2}$

$\Rightarrow c^{2}+a^{2} \tan ^{2} x-2 a c \tan x=b^{2} \sec ^{2} x$

$\Rightarrow c^{2}+a^{2} \tan ^{2} x-2 a c \tan x=b^{2}\left(1+\tan ^{2} x\right)$

$\Rightarrow\left(a^{2}-b^{2}\right) \tan ^{2} x-2 a c \tan x+\left(c^{2}-b^{2}\right)=0$

This is a quadratic in $\tan x$.

It has two solutions $\tan \alpha$ and $\tan \beta$.

$\tan \alpha+\tan \beta=\frac{2 a c}{a^{2}-b^{2}}$

$\tan \alpha \times \tan \beta=\frac{c^{2}-b^{2}}{a^{2}-b^{2}}$

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

$=\frac{\frac{2 a c}{a^{2}-b^{2}}}{1-\frac{c^{2}-b^{2}}{a^{2}-b^{2}}}$

$=\frac{2 a c}{a^{2}-c^{2}}$

Hence, $\sin (\alpha+\beta)=\frac{2 a c}{a^{2}+c^{2}}$ and $\cos (\alpha+\beta)=\frac{a^{2}-c^{2}}{a^{2}+c^{2}}$.