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यदि $\left| {\cos \,\theta \,\left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}\,} \right|\, \le k,$ तब $k$ का मान है
$\sqrt {1 + {{\cos }^2}\alpha } $
$\sqrt {1 + {{\sin }^2}\alpha } $
$\sqrt {2 + {{\sin }^2}\alpha } $
$\sqrt {2 + {{\cos }^2}\alpha } $
Solution
माना $u = \cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}$
$ \Rightarrow {(u – \sin \theta \cos \theta )^2} = {\cos ^2}\theta ({\sin ^2}\theta + {\sin ^2}\alpha )$
$ \Rightarrow {u^2}{\tan ^2}\theta – 2u\tan \theta + {u^2} – {\sin ^2}\alpha = 0$
चूँकि $\tan \theta $ वास्तविक है।
अत: $4{u^2} – 4{u^2}({u^2} – {\sin ^2}\alpha ) \ge 0$
$ \Rightarrow {u^2} – (1 + {\sin ^2}\alpha ) \le 0$
$ \Rightarrow |u|\, \le \sqrt {1 + {{\sin }^2}\alpha } $.
