If f(x)=(cosx)(cos2x)−−−−(cosnx) then f1x+∑r=1nrTanrxfx=

f(x)=cosx.cos2x.cos3x−−−−−cosnx logf(x)=logcosx+logcos2x+−−−−+log cosnx f'(x)f(x)=−(Tanx+2Tan2x+−−−−+nTan nx) =− ∑r=1nrtanrxf'(x)=−∑r=1n(rTanrx)f(x)f'(x)+∑r=1n(rTanrx)f(x)=0

If f(x)=(cosx)(cos2x)−−−−(cosnx) then f1x+∑r=1nrTanrxfx=

f(x)=cosx.cos2x.cos3x−−−−−cosnx logf(x)=logcosx+logcos2x+−−−−+log cosnx f'(x)f(x)=−(Tanx+2Tan2x+−−−−+nTan nx) =− ∑r=1nrtanrxf'(x)=−∑r=1n(rTanrx)f(x)f'(x)+∑r=1n(rTanrx)f(x)=0

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If $f (x) = (\cos x) (\cos 2 x) - - - - (\cos n x)$ then $f^{1} (x) + \sum_{r = 1}^{n} (r Tan r x) f (x) =$

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$f (x)$

$- f (x)$

$2 f (x)$

answer is B.

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Detailed Solution

$f (x) = \cos x . \cos 2 x . \cos 3 x - - - - - \cos n x \log f (x) = logcos x + logcos 2 x + - - - - + \log \cos n x \frac{f^{'} (x)}{f (x)} = - (T a n x + 2 T a n 2 x + - - - - + n T a n n x) = -_{} \sum_{r = 1}^{n} r \tan r x$