If the point $(1-\cos \theta ,\sin \theta )$  lies between the region corresponding to acute angle between the lines x – y = 0 and 6y = x , then

$P(1-\cos \theta ,\sin \theta )$                                                                                                            $(1-\cos \theta -\sin \theta )(1-\cos \theta -6\sin \theta )<0$                                                                                                  $\left(2{\sin }^2\frac{\theta }{2}-2\sin \frac{\theta }{2}\right)$                                                                                                                                $\left(2{\sin }^2\frac{\theta }{2}-6\times 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right)<0$                                                                                      $4{\sin }^2\frac{\theta }{2}\left(\sin \frac{\theta }{2}-\cos \frac{\theta }{2}\right)\left(\sin \frac{\theta }{2}-6\cos \frac{\theta }{2}\right)<0$                                                                          $4{\sin }^2\frac{\theta }{2}{\cos }^2\frac{\theta }{2}\left(\tan \frac{\theta }{2}-1\right)\left(\tan \frac{\theta }{2}-6\right)<0$                                                                                             $\left(\tan \frac{\theta }{2}-1\right)\left(\tan \frac{\theta }{2}-6\right)<0$                                                                                                                                                                                                 $1<\tan \frac{\theta }{2}<6$                                                                                                                                 $n\pi +{\tan }^{-1}1<\frac{\theta }{2}<n\pi +{\tan }^{-1}6$                                                                                                   $\Rightarrow 2n\pi +\frac{\pi }{2}<\theta <2n\pi +2{\tan }^{-1}6$