Question

Let $a,b$ be the roots of the equation $x^2-x+\sin 2\theta \cos \theta =x\sin \theta (2-\sin \theta )$ where $\frac{\pi }{4}<\theta <\frac{\pi }{2}$ and a>b. Find the value of $a+\frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}$

• A. $\sin \theta +1$
• B. $\cos \theta +1$
• C. $\sin \theta -1$
• D. $\cos \theta -1$

Answer 1

The correct option is A $\sin \theta +1$

Let
$k=a+\frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}$
$\Rightarrow k=a+\frac{b}{k}$
$\Rightarrow k^2-ak-b=0$
$\Rightarrow k=\frac{a\pm \sqrt{a^2+4b}}{2}$

$x^2-x+\sin 2\theta \cos \theta =x\sin \theta (2-\sin \theta )$
$\Rightarrow x^2-x(2\sin \theta +{\cos }^2\theta )+2\sin \theta {\cos }^2\theta =0\Rightarrow (x-2\sin \theta )(x-{\cos }^2\theta )=0$
$\Rightarrow x=2\sin \theta ,{\cos }^2\theta$
When $\frac{\pi }{4}<\theta <\frac{\pi }{2}$, ${\cos }^2\theta <2\sin \theta$
So,
$a=2\sin \theta$ and $b={\cos }^2\theta$
$k=\frac{2\sin \theta \pm \sqrt{4{\sin }^2\theta +4{\cos }^2\theta }}{2}=\frac{2\sin \theta \pm 2}{2}=\sin \theta \pm 1$
As $a,b$ is always positive in the given range, so
$k=\sin \theta +1$