Question

If $\left|\cos \theta \{\sin \theta +\sqrt{{\sin }^2\theta +{\sin }^2\alpha }\}\right|\le k,$ 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

Answer: (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,therefore

$4u^2-4u^2(u^2-{\sin }^2\alpha )\ge 0\Rightarrow u^2-(1+{\sin }^2\alpha )\le 0\Rightarrow \left|u\right|\le \sqrt{1+{\sin }^2\alpha }$