50 tan left(3 tan ^{-1}left(frac{1}{2}right)+ | vedclass

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(A) Let $A = 3 	an ^{-1} \frac{1}{2} + 2 \cos ^{-1} \frac{1}{\sqrt{5}}$.Since $\cos ^{-1} \frac{1}{\sqrt{5}} = 	an ^{-1} 2$,we have $A = 	an ^{-1}

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(A) Let $A = 3 	an ^{-1} \frac{1}{2} + 2 \cos ^{-1} \frac{1}{\sqrt{5}}$.Since $\cos ^{-1} \frac{1}{\sqrt{5}} = 	an ^{-1} 2$,we have $A = 	an ^{-1} \frac{1}{2} + 2(	an ^{-1} \frac{1}{2} + 	an ^{-1} 2)$.Using $	an ^{-1} x + 	an ^{-1} y = \frac{\pi}{2}$ for $xy=1$,we get $A = 	an ^{-1} \frac{1}{2} + 2(\frac{\pi}{2}) = 	an ^{-1} \frac{1}{2} + \pi$.Thus,$	an A = 	an(	an ^{-1} \frac{1}{2} + \pi) = 	an(	an ^{-1} \frac{1}{2}) = \frac{1}{2}$.Now,let $B = \frac{1}{2} 	an ^{-1}(2 \sqrt{2})$. Let $	an ^{-1}(2 \sqrt{2}) = 	heta$,so $	an 	heta = 2 \sqrt{2}$.We need $	an(\frac{	heta}{2})$. Using $	an 	heta = \frac{2 	an(	heta/2)}{1 - 	an^2(	heta/2)}$,let $t = 	an(	heta/2)$.$2 \sqrt{2} = \frac{2t}{1 - t^2} \implies \sqrt{2} = \frac{t}{1 - t^2} \implies \sqrt{2} - \sqrt{2} t^2 = t \implies \sqrt{2} t^2 + t - \sqrt{2} = 0$.Solving for $t$: $t = \frac{-1 \pm \sqrt{1 - 4(\sqrt{2})(-\sqrt{2})}}{2 \sqrt{2}} = \frac{-1 \pm \sqrt{1 + 8}}{2 \sqrt{2}} = \frac{-1 \pm 3}{2 \sqrt{2}}$.Since $	an ^{-1}(2 \sqrt{2})$ is in $(0, \pi/2)$,$t > 0$,so $t = \frac{2}{2 \sqrt{2}} = \frac{1}{\sqrt{2}}$.The expression is $50(\frac{1}{2}) + 4 \sqrt{2}(\frac{1}{\sqrt{2}}) = 25 + 4 = 29$.

$50 \tan \left(3 \tan ^{-1}\left(\frac{1}{2}\right)+2 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1}(2 \sqrt{2})\right)$ is equal to

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