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Answer:  $\frac{2\pi }{3}$
Hint: The range of the principal value of   $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given:    $\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )$
Solution:  We know that $\tan \left ( \frac{3\pi }{4} \right )= -1$
$\therefore 2\tan \left ( \frac{3\pi }{4} \right )\Rightarrow 2\times -1$
$2\tan \left ( \frac{3\pi }{4} \right )\Rightarrow -2$
By substituting these values in$\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )$,we get
$\sec ^{-1}\left ( -2 \right )$
Let ,$y= \sec ^{-1}\left ( -2 \right )$
$\sec \left ( y \right )= -2$
$-\sec\left ( \frac{\pi }{3} \right )= -2$
$\Rightarrow \sec\left ( \pi - \frac{\pi }{3} \right )$
$\Rightarrow \sec\left ( \frac{2\pi }{3} \right )$
$\therefore$The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and\, \sec \left ( \frac{2\pi }{3} \right )\! =\! -2$
$\therefore$The principal value of $\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )\, is\, \frac{2\pi }{3}$

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