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$\frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} = z$

${\mathop{\rm Im}\nolimits} (z) = 0$

$\theta = ?$

$z = \frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} \times \frac{{1 - i\cos \theta }}{{1 - i\cos \theta }}$

$ = \frac{{1 - i(\cos \theta  + \sin \theta ) + {i^2}\sin \theta \cos \theta }}{{1 - {i^2}{{\cos }^2}\theta }}$

$ = \frac{{1 - \sin \theta \cos \theta  - i(\cos \theta  + \sin \theta )}}{{1 + {{\cos }^2}\theta }}$

$\because {\mathop{\rm Im}\nolimits} (z) = 0,\cos \theta  + \sin \theta  = 0$

$\Rightarrow \tan \theta = -1 $

$\Rightarrow \theta = n\pi - \frac {\pi}{4}$,$n\in I$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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