tan left(2 tan ^{-1} frac{1}{5} + sec ^{-1} f | vedclass

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(B) Let the expression be $E = 	an \left(2 	an ^{-1} \frac{1}{5} + 2 	an ^{-1} \frac{1}{8} + \sec ^{-1} \frac{\sqrt{5}}{2}ight)$.First,simplify $

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$2$

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(B) Let the expression be $E = 	an \left(2 	an ^{-1} \frac{1}{5} + 2 	an ^{-1} \frac{1}{8} + \sec ^{-1} \frac{\sqrt{5}}{2}ight)$.First,simplify $\sec ^{-1} \frac{\sqrt{5}}{2}$. Let $	heta = \sec ^{-1} \frac{\sqrt{5}}{2}$,then $\sec 	heta = \frac{\sqrt{5}}{2}$,so $\cos 	heta = \frac{2}{\sqrt{5}}$. Thus,$	an 	heta = \sqrt{(\sqrt{5}/2)^2 - 1} = \sqrt{5/4 - 1} = \sqrt{1/4} = \frac{1}{2}$. So,$\sec ^{-1} \frac{\sqrt{5}}{2} = 	an ^{-1} \frac{1}{2}$.Now,$E = 	an \left(2 \left(	an ^{-1} \frac{1}{5} + 	an ^{-1} \frac{1}{8}ight) + 	an ^{-1} \frac{1}{2}ight)$.Using $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$,we get $	an ^{-1} \frac{1}{5} + 	an ^{-1} \frac{1}{8} = 	an ^{-1} \left(\frac{1/5 + 1/8}{1 - 1/40}ight) = 	an ^{-1} \left(\frac{13/40}{39/40}ight) = 	an ^{-1} \frac{1}{3}$.So,$E = 	an \left(2 	an ^{-1} \frac{1}{3} + 	an ^{-1} \frac{1}{2}ight)$.Using $2 	an ^{-1} x = 	an ^{-1} \left(\frac{2x}{1-x^2}ight)$,we get $2 	an ^{-1} \frac{1}{3} = 	an ^{-1} \left(\frac{2/3}{1-1/9}ight) = 	an ^{-1} \left(\frac{2/3}{8/9}ight) = 	an ^{-1} \left(\frac{2}{3} 	imes \frac{9}{8}ight) = 	an ^{-1} \frac{3}{4}$.Now,$E = 	an \left(	an ^{-1} \frac{3}{4} + 	an ^{-1} \frac{1}{2}ight) = 	an \left(	an ^{-1} \left(\frac{3/4 + 1/2}{1 - (3/4)(1/2)}ight)ight) = \frac{5/4}{1 - 3/8} = \frac{5/4}{5/8} = \frac{5}{4} 	imes \frac{8}{5} = 2$.

$\tan \left(2 \tan ^{-1} \frac{1}{5} + \sec ^{-1} \frac{\sqrt{5}}{2} + 2 \tan ^{-1} \frac{1}{8}\right)$ is equal to.

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