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Prove that, ${\left( {\frac{{1 + \sin \theta + i\cos \theta }}{{1 + \sin \theta - i\cos \theta }}} \right)^n} =$

$\cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

LHS = ${\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{\{ 1 + (\sin \theta  - i\cos \theta )\} (\sin \theta  + i\cos \theta )}}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(\sin \theta  + i\cos \theta ) + \{ {{\sin }^2}\theta  - {{(i\cos \theta )}^2}\} }}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(\sin \theta  + i\cos \theta ) + ({{\sin }^2}\theta  + {{\cos }^2}\theta )}}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(1 + \sin \theta  + i\cos \theta )}}} \right]^n}$

$ = {(\sin \theta  + i\cos \theta )^n}$

$ = {\left\{ {\cos \left( {\frac{\pi }{2} - \theta } \right) + i\sin \left( {\frac{\pi }{2} - \theta } \right)} \right\}^n}$

$ = \cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

= RHS

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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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