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Question
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
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Solution
Given: \[a \cos2x + b \sin2x = c\]
\[\Rightarrow a\left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + b\left( \frac{2\text{ tan } x}{1 + \tan^2 x} \right) - c = 0\]
\[ \Rightarrow a\left( 1 - \tan^2 x \right) + 2b \text{ tan } x - c\left( 1 + \tan^2 x \right) = 0\]
\[ \Rightarrow a - a \tan^2 x + 2b \text{ tan } x - c - c \tan^2 x = 0\]
\[ \Rightarrow \left( a + c \right) \tan^2 x - 2b \text{ tan } x + \left( c - a \right) = 0 . . . . . \left( 1 \right)\]
This a quadratic equation in terms of \[\tan^2 x\] .
It is given that α and β are the roots of the given equation, so tan α and tan β are the roots of (1).
Since tan α and tan β are the roots of the equation
\[ = \frac{\frac{2b}{a + c}}{1 - \left( \frac{c - a}{c + a} \right)} \left[ \text{ From } \left( i \right) \text{ and } \left( ii \right) \right]\]
\[ = \frac{\frac{2b}{a + c}}{\frac{c + a - c + a}{c + a}}\]
\[ = \frac{2b}{2a}\]
\[ = \frac{b}{a}\]
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