If  $f\left(x\right)=\left(\cos x\right)\left(\cos 2x\right)----\left(\cos nx\right)$  then  $f^1\left(x\right)+\sum _{r=1}^n\left(r\mathrm{Tan}rx\right)f\left(x\right)=$

$$\begin{gathered} f\left(x\right)=\cos x.\cos 2x.\cos 3x-----\cos nx \\ \log f\left(x\right)=\mathrm{logcos}x+\mathrm{logcos}2x+----+\log \cos nx \\ \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=-(Tanx+2Tan2x+----+nTannx) \\ ={-}_{ }\sum _{r=1}^{n}r\tan rx \end{gathered}$$

$\begin{array}{l}{f}^{\text{'}}\left(x\right)=-\sum _{r=1}^n\left(rTanrx\right)f\left(x\right)\\ f^{\text{'}}\left(x\right)+\sum _{r=1}^n\left(rTanrx\right)f\left(x\right)=0\end{array}$