$50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$ $+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)$ $\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$ $\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$ Where $2 \alpha=\tan ^{-1} 2 \sqrt{2}$ $\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad$.. (i) $\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0$ $\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0$ $\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0$ $\Rightarrow \tan \alpha=\sqrt{2}$ or $\frac{1}{\sqrt{2}}$ $\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}$ $(\tan \alpha=\sqrt{2}$ doesn't satisfy (i)) $\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$