2 tan ^{-1} frac{1}{5}+sec ^{-1} frac{5 sqrt{ | vedclass

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(B) Given that, $2 	an ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 	an ^{-1} \frac{1}{8}$ $=2 \left(	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{

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$\frac{\pi}{4}$

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(B) Given that, $2 	an ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 	an ^{-1} \frac{1}{8}$ $=2 \left(	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{8}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$ Using the formula $	an ^{-1} A+	an ^{-1} B=	an ^{-1}\left(\frac{A+B}{1-A B}ight)$, we get: $=2 	an ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$ $=2 	an ^{-1}\left(\frac{13}{39}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = 2 	an ^{-1}\left(\frac{1}{3}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$ Using $2 	an ^{-1} A=	an ^{-1}\left(\frac{2 A}{1-A^2}ight)$, we get: $=	an ^{-1}\left(\frac{2 	imes \frac{1}{3}}{1-\frac{1}{9}}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = 	an ^{-1}\left(\frac{2/3}{8/9}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = 	an ^{-1}\left(\frac{3}{4}ight)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$ Now, convert $\sec ^{-1} \frac{5 \sqrt{2}}{7}$ to $	an ^{-1}$. Let $	heta = \sec ^{-1} \frac{5 \sqrt{2}}{7}$, then $\sec 	heta = \frac{5 \sqrt{2}}{7}$. The opposite side is $\sqrt{(5 \sqrt{2})^2 - 7^2} = \sqrt{50 - 49} = 1$. Thus, $	an 	heta = \frac{1}{7}$, so $\sec ^{-1} \frac{5 \sqrt{2}}{7} = 	an ^{-1} \frac{1}{7}$. The expression becomes $	an ^{-1}\left(\frac{3}{4}ight)+	an ^{-1}\left(\frac{1}{7}ight)$ $= 	an ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}ight) = 	an ^{-1}\left(\frac{21+4}{28-3}ight) = 	an ^{-1}\left(\frac{25}{25}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$

$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}=$

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