If \[ a\cos2x+b\sin2x=c \] has roots \(\alpha\) and \(\beta\), prove that
\[ (i)\quad \tan\alpha+\tan\beta=\frac{2b}{a+c} \]
\[ (ii)\quad \tan\alpha\tan\beta=\frac{c-a}{c+a} \]
\[ (iii)\quad \tan(\alpha+\beta)=\frac{b}{a} \]
Question
If \[ a\cos2x+b\sin2x=c \] has \(\alpha\) and \(\beta\) as its roots, prove that
\[ (i)\quad \tan\alpha+\tan\beta=\frac{2b}{a+c} \]
\[ (ii)\quad \tan\alpha\tan\beta=\frac{c-a}{c+a} \]
\[ (iii)\quad \tan(\alpha+\beta)=\frac{b}{a} \]
Solution
Given,
\[ a\cos2x+b\sin2x=c \]
Using
\[ \cos2x=\frac{1-\tan^2x}{1+\tan^2x} \]
and
\[ \sin2x=\frac{2\tan x}{1+\tan^2x} \]
Substituting,
\[ a\left( \frac{1-\tan^2x}{1+\tan^2x} \right) + b\left( \frac{2\tan x}{1+\tan^2x} \right) = c \]
Multiplying throughout by \[ 1+\tan^2x \]
\[ a(1-\tan^2x)+2b\tan x = c(1+\tan^2x) \]
\[ a-a\tan^2x+2b\tan x = c+c\tan^2x \]
Rearranging,
\[ (a+c)\tan^2x-2b\tan x+(c-a)=0 \]
This is a quadratic equation in \[ \tan x \]
Since \(\alpha\) and \(\beta\) are roots,
\[ \tan\alpha \quad \text{and} \quad \tan\beta \]
are roots of
\[ (a+c)t^2-2bt+(c-a)=0 \]
Using the sum of roots formula,
\[ \tan\alpha+\tan\beta = \frac{-(-2b)}{a+c} \]
\[ \boxed{ \tan\alpha+\tan\beta = \frac{2b}{a+c} } \]
Using the product of roots formula,
\[ \tan\alpha\tan\beta = \frac{c-a}{a+c} \]
\[ \boxed{ \tan\alpha\tan\beta = \frac{c-a}{c+a} } \]
Now,
\[ \tan(\alpha+\beta) = \frac{ \tan\alpha+\tan\beta }{ 1-\tan\alpha\tan\beta } \]
Substituting the obtained values,
\[ = \frac{ \frac{2b}{a+c} }{ 1-\frac{c-a}{c+a} } \]
\[ = \frac{ \frac{2b}{a+c} }{ \frac{(a+c)-(c-a)}{a+c} } \]
\[ = \frac{ \frac{2b}{a+c} }{ \frac{2a}{a+c} } \]
\[ = \frac{b}{a} \]
Therefore,
\[ \boxed{ \tan(\alpha+\beta)=\frac{b}{a} } \]
Hence proved.