50 tan(3 tan^-1( 1 2)+2 cos^-1( 1 √5))+4√2 tan( 1 2 tan^-1(2√2)) is equal to .

Let $S=50\mathrm{tan}(3{\mathrm{tan}}^{-1}\frac{1}{2}+2{\mathrm{cos}}^{-1}\frac{1}{\sqrt{5}})$S=50tan(3tan−121​+2cos−15​1​)$+4\sqrt{2}\mathrm{tan}(\frac{1}{2}{\mathrm{tan}}^{-1}2\sqrt{2})$+42​tan(21​tan−122​)

$=50\mathrm{tan}({\mathrm{tan}}^{-1}\frac{1}{2}+2({\mathrm{tan}}^{-1}\frac{1}{2}+{\mathrm{tan}}^{-1}2))$=50tan(tan−121​+2(tan−121​+tan−12))$+4\sqrt{2}\mathrm{tan}(\frac{1}{2}{\mathrm{tan}}^{-1}2\sqrt{2})$+42​tan(21​tan−122​)

Let $\frac{1}{2}{\mathrm{tan}}^{-1}2\sqrt{2}=\alpha \Rightarrow \mathrm{tan}2\alpha =2\sqrt{2}$21​tan−122​=α⇒tan2α=22​

Now $\frac{2\mathrm{tan}\alpha }{1-{\mathrm{tan}}^{2}\alpha }=2\sqrt{2}$1−tan2α2tanα​=22​

Solving we get, $\mathrm{tan}\alpha =\frac{1}{\sqrt{2}}$tanα=2​1​

So$S=50\mathrm{tan}({\mathrm{tan}}^{-1}\frac{1}{2}+2\cdot \frac{\pi }{2}))+4\sqrt{2}\times \frac{1}{\sqrt{2}}$S=50tan(tan−121​+2⋅2π​))+42​×2​1​

$$=50({\mathrm{tantan}}^{-1}\frac{1}{2})+4$$=50(tantan−121​)+4

$$=25+4=29$$=25+4=29