If Tan−1⁡15+12Sec−1⁡x+Tan−1⁡18=π8, then x2=

Tan−1⁡15+12Sec−1⁡x+Tan−1⁡18=π8⇒12Sec−1⁡x+Tan−1⁡15+181−1518=π8⇒12Sec−1⁡x+Tan−1⁡1339=π8⇒Sec−1⁡x+2Tan−1⁡13=π4⇒Sec−1⁡x+Tan−1⁡2(1/3)1−1/9=π4⇒Sec−1⁡x+Tan−1⁡34=π4⇒Sec−1⁡x=Tan−1⁡1−Tan−1⁡34=Tan−1⁡1−3/41+3/4=Tan−1⁡17=Sec−1⁡1+497⇒x=507⇒x2=5049.

If Tan−1⁡15+12Sec−1⁡x+Tan−1⁡18=π8, then x2=

Tan−1⁡15+12Sec−1⁡x+Tan−1⁡18=π8⇒12Sec−1⁡x+Tan−1⁡15+181−1518=π8⇒12Sec−1⁡x+Tan−1⁡1339=π8⇒Sec−1⁡x+2Tan−1⁡13=π4⇒Sec−1⁡x+Tan−1⁡2(1/3)1−1/9=π4⇒Sec−1⁡x+Tan−1⁡34=π4⇒Sec−1⁡x=Tan−1⁡1−Tan−1⁡34=Tan−1⁡1−3/41+3/4=Tan−1⁡17=Sec−1⁡1+497⇒x=507⇒x2=5049.

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If ${Tan}^{- 1} \frac{1}{5} + \frac{1}{2} {Sec}^{- 1} x + {Tan}^{- 1} \frac{1}{8} = \frac{π}{8}, then x^{2}$ =

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$\frac{13}{12}$

$\frac{50}{49}$

$\frac{12}{7}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

${Tan}^{- 1} \frac{1}{5} + \frac{1}{2} {Sec}^{- 1} x + {Tan}^{- 1} \frac{1}{8} = \frac{π}{8} \Rightarrow \frac{1}{2} {Sec}^{- 1} x + {Tan}^{- 1} \frac{\frac{1}{5} + \frac{1}{8}}{1 - (\frac{1}{5}) (\frac{1}{8})} = \frac{π}{8}$
$\begin{array}{l} \Rightarrow \frac{1}{2} {Sec}^{- 1} x + {Tan}^{- 1} \frac{13}{39} = \frac{π}{8} \Rightarrow {Sec}^{- 1} x + 2 {Tan}^{- 1} \frac{1}{3} = \frac{π}{4} \\ \Rightarrow {Sec}^{- 1} x + {Tan}^{- 1} (\frac{2 (1 / 3)}{1 - 1 / 9}) = \frac{π}{4} \Rightarrow {Sec}^{- 1} x + {Tan}^{- 1} (\frac{3}{4}) = \frac{π}{4} \\ \Rightarrow {Sec}^{- 1} x = {Tan}^{- 1} 1 - {Tan}^{- 1} \frac{3}{4} = {Tan}^{- 1} (\frac{1 - 3 / 4}{1 + 3 / 4}) = {Tan}^{- 1} (\frac{1}{7}) = {Sec}^{- 1} (\frac{\sqrt{1 + 49}}{7}) \\ \Rightarrow x = \frac{\sqrt{50}}{7} \Rightarrow x^{2} = \frac{50}{49} . \end{array}$