If left| {cos ,theta ,left{ {sin theta + sqrt | vedclass

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Question

(b) Let $u = \cos 	heta \left\{ {\sin 	heta + \sqrt {{{\sin }^2}	heta + {{\sin }^2}\alpha } } ight\}$

==> ${(u - \sin 	heta \cos 	heta )^

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Let $u = \cos 	heta \left\{ {\sin 	heta + \sqrt {{{\sin }^2}	heta + {{\sin }^2}\alpha } } ight\}$

==> ${(u - \sin 	heta \cos 	heta )^2} = {\cos ^2}	heta ({\sin ^2}	heta + {\sin ^2}\alpha )$ 

==> ${u^2}{	an ^2}	heta - 2u	an 	heta + {u^2} - {\sin ^2}\alpha = 0$ 

Since tan $	heta $ is real, therefore 

==> $4{u^2} - 4{u^2}({u^2} - {\sin ^2}\alpha ) \ge 0$

$ \Rightarrow {u^2} - (1 + {\sin ^2}\alpha ) \le 0$ 

==> $|u|\, \le \sqrt {1 + {{\sin }^2}\alpha } $.

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

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