Question

If $\alpha$ and $\beta$ are the roots of the equation a $\cos 2\theta +b\sin 2\theta =c$ then ${\cos }^2\alpha +{\cos }^2\beta$ is equal to

• A. $\frac{a^2+ac+b^2}{a^2+b^2}$
• B. $\frac{a^2-ac+b^2}{a^2+b^2}$
• C. $\frac{2b^2}{a^2+c^2}$
• D. $\frac{2a^2}{b^2+c^2}$

Answer 1

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

Given $a\cos 2\theta +b\sin 2\theta =c$
$\Rightarrow b\sin 2\theta =c-a\cos 2\theta$
squaring both side
$\Rightarrow b^2{\sin }^{2}2\theta =c^2-2ac\cos 2\theta +a^2{\cos }^{2}2\theta$
$\Rightarrow b^2-b^2{\cos }^{2}2\theta =c^2-2ac\cos 2\theta +a^2{\cos }^{2}2\theta$
$\Rightarrow (a^2+b^2){\cos }^{2}2\theta -2ac\cos 2\theta +(c^2-b^2)=0$
$\Rightarrow \cos 2\theta =\frac{2ac\pm \sqrt{4a^2{c}^2-4(a^2+b^2)(c^2-b^2)}}{2(a^2+b^2)}$
$\Rightarrow 2{\cos }^2\theta -1=\frac{ac\pm \sqrt{a^2{b}^2-b^2{c}^2+b^4}}{(a^2+b^2)}$
$\Rightarrow {\cos }^2\theta =\frac{a^2+b^2+ac\pm \sqrt{a^2{b}^2-b^2{c}^2+b^4}}{2(a^2+b^2)}$
also given $\alpha$ and $\beta$ are the roots, so
${\cos }^2\alpha +{\cos }^2\beta =\frac{a^2+b^2+ac}{a^2+b^2}$
Hence, option 'A' is correct.