Question

If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that 

(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]

 

Answer 1

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

 \[\left( a + c \right) \tan^2 x - 2b \text{ tan } x + \left( c - a \right) = 0\]. Therefore,

(i) \[\tan\alpha + \tan\beta = - \frac{\left( - 2b \right)}{a + c} \left( \text{ Sum of roots } = - \frac{b}{a} \right)\] 
\[ \Rightarrow \tan\alpha + \tan\beta = \frac{2b}{a + c}\]