Question

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

Answer 1

Let$u=cos\theta \{sin\theta +\sqrt{si{n}^2\theta +si{n}^2\alpha }\}\Rightarrow (u-sin\theta cos\theta {)}^2=co{s}^2\theta (si{n}^2\theta +si{n}^2\alpha )\Rightarrow u^2+si{n}^2\theta co{s}^2\theta -2usin\theta cos\theta =co{s}^2\theta (si{n}^2\theta +si{n}^2\alpha )\Rightarrow u^{2}se{c}^2\theta +si{n}^2\theta -2utan\theta =si{n}^2\theta +si{n}^2\alpha \Rightarrow u^{2}ta{n}^2\theta -2utan\theta +u^2-si{n}^2\alpha =0$

Since tan $\theta$ is real, therefore

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