If |cos theta {sin theta + sqrt{sin^2 theta + | vedclass

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(B) Let $u = \cos 	heta \{\sin 	heta + \sqrt{\sin^2 	heta + \sin^2 \alpha\}}$.Rearranging the terms,we have $u - \sin 	heta \cos 	heta = \cos 	h

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

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(B) Let $u = \cos 	heta \{\sin 	heta + \sqrt{\sin^2 	heta + \sin^2 \alpha\}}$.Rearranging the terms,we have $u - \sin 	heta \cos 	heta = \cos 	heta \sqrt{\sin^2 	heta + \sin^2 \alpha}$.Squaring both sides:$(u - \sin 	heta \cos 	heta)^2 = \cos^2 	heta (\sin^2 	heta + \sin^2 \alpha)$.$u^2 - 2u \sin 	heta \cos 	heta + \sin^2 	heta \cos^2 	heta = \sin^2 	heta \cos^2 	heta + \cos^2 	heta \sin^2 \alpha$.$u^2 - 2u \sin 	heta \cos 	heta - \cos^2 	heta \sin^2 \alpha = 0$.Dividing by $\cos^2 	heta$ (assuming $\cos 	heta 
eq 0$):$u^2 \sec^2 	heta - 2u 	an 	heta - \sin^2 \alpha = 0$.Using $\sec^2 	heta = 1 + 	an^2 	heta$:$u^2 (1 + 	an^2 	heta) - 2u 	an 	heta - \sin^2 \alpha = 0$.$u^2 	an^2 	heta - 2u 	an 	heta + (u^2 - \sin^2 \alpha) = 0$.Since $	an 	heta$ is real,the discriminant $D \ge 0$:$(-2u)^2 - 4(u^2)(u^2 - \sin^2 \alpha) \ge 0$.$4u^2 - 4u^4 + 4u^2 \sin^2 \alpha \ge 0$.Dividing by $4u^2$ (assuming $u 
eq 0$):$1 - u^2 + \sin^2 \alpha \ge 0$.$u^2 \le 1 + \sin^2 \alpha$.Therefore,$|u| \le \sqrt{1 + \sin^2 \alpha}$.Thus,$k = \sqrt{1 + \sin^2 \alpha}$.

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

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