Question

If $|\cos \theta \{\sin \theta +\sqrt{{\sin }^2\theta +{\sin }^2\alpha }\}|\le k$, then the value of $k$ is

• A. $\sqrt{1+{\cos }^2\alpha }$
• B. $\sqrt{1+{\sin }^2\alpha }$
• C. $\sqrt{2+{\sin }^2\alpha }$
• D. $\sqrt{2+{\cos }^2\alpha }$

Answer 1

The correct option is B $\sqrt{1+{\sin }^2\alpha }$

Let $u=\cos \theta \{\sin \theta +\sqrt{{\sin }^2\theta +{\sin }^2\alpha }\}$
$\Rightarrow (u-\sin \theta \cos \theta {)}^2={\cos }^2\theta ({\sin }^2\theta +{\sin }^2\alpha )$
$\Rightarrow u^2{\tan }^2\theta -2u\tan \theta +u^2-{\sin }^2\alpha =0$

Since $\tan \theta$ is real, we have
$D\ge 0$
$\Rightarrow 4u^2-4u^2(u^2-{\sin }^2\alpha )\ge 0$
$\Rightarrow u^2\le 1+{\sin }^2\alpha$
$\Rightarrow |u|\le \sqrt{1+{\sin }^2\alpha }$