3.Trigonometrical Ratios, Functions and Identities
hard

यदि $\left| {\cos \,\theta \,\left\{ {\sin \theta  + \sqrt {{{\sin }^2}\theta  + {{\sin }^2}\alpha } } \right\}\,} \right|\, \le k,$ तब $k$ का मान है

A

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

B

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

C

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

D

$\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 } $.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.