Question

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

The correct option is B

$\sqrt{(1+{\sin }^2\alpha )}$

Explanation for the correct option:

Step 1. Find the value of $k$:

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

Let $p=cos\theta \left\{\sin \theta +\sqrt{({\sin }^2\theta +{\sin }^2\alpha )}\right\}$ ….(i)

$\Rightarrow$ $p=\cos \theta \sin \theta +\cos \theta \sqrt{({\sin }^2\theta +{\sin }^2\alpha )}$

$\Rightarrow$$p–\cos \theta \sin \theta =\cos \theta \sqrt{({\sin }^2\theta +{\sin }^2\alpha )}$

Step 2. Square above equation both sides, we get

$p^2+{\cos }^2\theta {\sin }^2\theta –2p\cos \theta \sin \theta ={\cos }^2\theta ({\sin }^2\theta +{\sin }^2\alpha )$

$\Rightarrow$$p^2+{\cos }^2\theta {\sin }^2\theta –2p\cos \theta \sin \theta ={\cos }^2\theta {\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha$

$\Rightarrow$ $p^2–2p\sin \theta \cos \theta ={\cos }^2\theta {\sin }^2\alpha$

Step 3. Divide above equation by ${\cos }^2\theta$on both sides, we get

$p^{2}se{c}^2\theta –2p\tan \theta ={\sin }^2\alpha$

$\Rightarrow$ $p^2(1+{\tan }^2\theta )–2p\tan \theta ={\sin }^2\alpha$

$\Rightarrow$ $p^2+p^2{\tan }^2\theta –2p\tan \theta ={\sin }^2\alpha$

$\Rightarrow$ $p^2{\tan }^2\theta –2p\tan \theta +(p^2–{\sin }^2\alpha )=0$

It is a quadratic equation in $\tan \theta$.

If $\tan \theta$ is real, then

${(-2p)}^2–4\left(p^2\right)(p^2–{\sin }^2\alpha )\ge 0$

$\Rightarrow$ $4p^2[1–p^2+{\sin }^2\alpha ]\ge 0$

$\Rightarrow$ $1–p^2+{\sin }^2\alpha \ge 0$

$\Rightarrow$ $p^2\le 1+{\sin }^2\alpha$ ….(ii)

Step 4. By equating equation (i) and (ii), we get

$\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\alpha }\mathbf{)}}$

Hence, Option ‘B’ is Correct.