Question

If $\alpha$ & $\beta$ satisfy the equation, $a\cos 2\theta +b\sin 2\theta =c$ then ${\cos }^2\alpha +{\cos }^2\beta =$

• A. $\frac{a^2+ac+b^2}{a^2+b^2}$.
• B. $\frac{a^2+c+b^2}{a^2+b^2}$.
  
• C. $\frac{a^2+a+b^2}{a^2+b^2}$.
  
• D. None of the above

Answer 1

The correct option is A $\frac{a^2+ac+b^2}{a^2+b^2}$.

$a\cos 2\theta +b\sin 2\theta =c$
$\Rightarrow a\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+b\frac{2\tan \theta }{1+{\tan }^2\theta }=c$
$\Rightarrow (c+a){\tan }^2\theta -\left(2b\right)\tan \theta +(c-a)$
Now if $\alpha ,\beta$ are the roots of this equation then,
$\tan \alpha +\tan \beta =\frac{2b}{c+a},\tan \alpha .\tan \beta =\frac{c-a}{c+a}$ (i)
$\therefore {\cos }^2\alpha +{\cos }^2\beta =\frac{1}{1+{\tan }^2\alpha }+\frac{1}{1+{\tan }^2\beta }$
$=\frac{{\tan }^2\alpha +{\tan }^2\beta +2}{1+{\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\alpha .{\tan }^{\beta }}$
$=\frac{(\tan \alpha +\tan \beta {)}^2-2\tan \alpha .\tan \beta +2}{1+(\tan \alpha +\tan \beta {)}^2-2\tan \alpha .\tan \beta +{\tan }^2\alpha .{\tan }^2\beta }$
$=\frac{a^2+ac+b^2}{a^2+b^2}$ using (i)