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

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