y=tan−1(1cos2x+cosx+1)+tan−1(1cos2x+3cosx+3)+....+ up to n terms then f'(0)=

y=tan−1((cosx+1)−cosx1+cosx(cosx+1))+tan−1((cosx+2)−(cosx+1)1+(cosx+1)(cosx+2))+........ tan−1((cosx+1)−(cosx+(n−1))1+(cosx+1)(cosx+(n−1)))y=tan−1(cosx+n)−tan−1(cosx)dydx=11+(cosx+n)2(−sinx)−11+cos2x(−sinx)⇒f'(0)=0

y=tan−1(1cos2x+cosx+1)+tan−1(1cos2x+3cosx+3)+....+ up to n terms then f'(0)=

y=tan−1((cosx+1)−cosx1+cosx(cosx+1))+tan−1((cosx+2)−(cosx+1)1+(cosx+1)(cosx+2))+........ tan−1((cosx+1)−(cosx+(n−1))1+(cosx+1)(cosx+(n−1)))y=tan−1(cosx+n)−tan−1(cosx)dydx=11+(cosx+n)2(−sinx)−11+cos2x(−sinx)⇒f'(0)=0

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$y = \tan^{- 1} (\frac{1}{\cos^{2} x + \cos x + 1}) + \tan^{- 1} (\frac{1}{\cos^{2} x + 3 \cos x + 3}) + .... +$ up to n terms then $f^{'} (0) =$

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Detailed Solution

$y = \tan^{- 1} (\frac{(\cos x + 1) - \cos x}{1 + \cos x (\cos x + 1)}) + \tan^{- 1} (\frac{(\cos x + 2) - (\cos x + 1)}{1 + (\cos x + 1) (\cos x + 2)}) + ........ \tan^{- 1} (\frac{(\cos x + 1) - (\cos x + (n - 1))}{1 + (\cos x + 1) (\cos x + (n - 1))})$