Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$Letu=cos\theta \{sin\theta +\sqrt{si{n}^2\theta +si{n}^2\alpha }\}\Rightarrow (u-sin\theta cos\theta {)}^2=co{s}^2\theta (si{n}^2\theta +si{n}^2\alpha )\Rightarrow u^{2}ta{n}^2\theta -2utan\theta +u^2-si{n}^2\alpha =0Sincetan\theta isreal,therefore\Rightarrow 4u^2-4u^2(u^2-si{n}^2\alpha )\ge 0\Rightarrow u^2-(1+si{n}^2\alpha )\le 0\Rightarrow |u|\le \sqrt{1+si{n}^2\alpha }$