The value of tan ^-1≤ft(tan (3 π)/4)+2 cos ^-1≤ft(cos (7 π)/6) is:

The correct answer is Option (1) → $17 \frac{\pi}{12}$

$\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)+2 \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$

$=\tan ^{-1}\left(-\tan \frac{\pi}{4}\right)+2 \cos ^{-1}\left(-\cos \frac{\pi}{6}\right)$

$=\tan ^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)=2 \cos ^{-1}\left(\cos \frac{5\pi}{6}\right)$

$=-\frac{\pi}{4}+2×\frac{5\pi}{6}=\frac{17\pi}{12}$