Correct Answer - Option 2 :
\( \frac {5\pi}{6} \)
Concept:
\(\arctan x = y ⇒ x = \tan y\)
Calculation:
Here, we have to find the value of \(\arctan \left( {1/\sqrt 3 } \right) + \operatorname{arcsec} \left( { - 2} \right)\)
\(\tan x = \frac{1}{{\sqrt 3 }}\) ⇒ \(x = \frac{\pi }{6}\)
Also, \(\sec y = -2\) ⇒ \(y = \frac{2\pi }{3}\)
So, \(\arctan \left( {1/\sqrt 3 } \right) + {\mathop{\rm arcsec}\nolimits} \left( { - 2} \right) = x + y\)
\(= \frac{\pi }{6} + \frac{2\pi }{3}\)
\(= \frac{\pi+4\pi}{6} = \frac {5\pi}{6} \)