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Question Number 78392 by jagoll last updated on 17/Jan/20

Answered by john santu last updated on 17/Jan/20

$$leth=\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}\sqrt[4]{\cos 4x}...\sqrt[n]{\cos nx}$$$$ln\left(h\right)=ln\left(\cos x\right)+\frac{1}{2}ln\left(\cos 2x\right)+\frac{1}{3}ln\left(\cos 3x\right)+...+\frac{1}{n}ln\left(\cos nx\right)$$$$\frac{dh}{dx}=\frac{-\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}-\frac{\sin 3x}{\cos 3x}-...-\frac{\sin nx}{\cos nx}$$$$nowweuse{L}^{\prime }Hopitalrule$$$$\underset{x\to 0}{\lim }\frac{\tan x+\tan 2x+\tan 3x+...+\tan nx}{2x}=$$$$\frac{1+2+3+4+...+n}{2}=\frac{n(n+1)}{4}$$

Commented by jagoll last updated on 17/Jan/20

$$thankssir$$

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