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

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(B) Given equation: $	an ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+	an ^{-1} \frac{1}{8}=\frac{\pi}{8}$ Using the formula $	an ^{-1} a + 	an ^{-1

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$\frac{50}{49}$

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(B) Given equation: $	an ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+	an ^{-1} \frac{1}{8}=\frac{\pi}{8}$ Using the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1} \left( \frac{a+b}{1-ab} ight)$, we get: $	an ^{-1} \left( \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}} ight) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$ $	an ^{-1} \left( \frac{\frac{13}{40}}{\frac{39}{40}} ight) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$ $	an ^{-1} \left( \frac{1}{3} ight) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$ Multiply by $2$: $2 	an ^{-1} \left( \frac{1}{3} ight) + \sec ^{-1} x = \frac{\pi}{4}$ Using $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$: $	an ^{-1} \left( \frac{2/3}{1-1/9} ight) + \sec ^{-1} x = \frac{\pi}{4}$ $	an ^{-1} \left( \frac{3}{4} ight) + \sec ^{-1} x = \frac{\pi}{4}$ Since $\sec ^{-1} x = 	an ^{-1} \sqrt{x^2-1}$, we have: $	an ^{-1} \left( \frac{3}{4} ight) + 	an ^{-1} \sqrt{x^2-1} = \frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{\frac{3}{4} + \sqrt{x^2-1}}{1 - \frac{3}{4} \sqrt{x^2-1}} = 	an \left( \frac{\pi}{4} ight) = 1$ $3 + 4 \sqrt{x^2-1} = 4 - 3 \sqrt{x^2-1}$ $7 \sqrt{x^2-1} = 1$ $\sqrt{x^2-1} = \frac{1}{7}$ $x^2 - 1 = \frac{1}{49}$ $x^2 = 1 + \frac{1}{49} = \frac{50}{49}$

List-$I$	List-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$

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