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$\begin{array}{*{20}{c}}{\frac{{\sin 0 + \sin 1 + \sin 2 + .......... + \sin n}}{{\cos 0 + \cos 1 + \cos 2 + ......... + \cos n}}}\\\parallel \\{\tan (?)}\end{array}$

L.H.S. $ = \frac{{\frac{{\sin \left( {0 + \frac{{n + 1 - 1}}{2}.1} \right)\sin \frac{{(n + 1).1}}{2}}}{{\sin \frac{1}{2}}}}}{{\frac{{\cos \left( {0 + \frac{{n + 1 - 1}}{2}.1} \right)\sin \frac{{(n + 1).1}}{2}}}{{\sin \frac{1}{2}}}}}$ 

(note that there are n+1 terms in both the numerator and the denominator)

$ = \frac{{\sin \frac{n}{2}}}{{\cos \frac{n}{2}}} = \tan \frac{n}{2}$

$\therefore \, ? = \frac{n}{2}$

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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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