Question

How do you evaluate $ {\sec ^{ - 1}}( - 2) $ ?

Answer 1

Hint: In order to find the solution of a trigonometric equation start by taking the inverse trig function (like inverse sin, inverse cosine, inverse tangent) of both sides of the equation and then set up reference angles to find the rest of the answers.
For $ {\sin ^{ - 1}} \to \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $
For $ {\cos ^{ - 1}} \to \left[ {0,\pi } \right] $
For $ {\tan ^{ - 1}}\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) $

Complete step-by-step answer:
According to definition of inverse ratio if $ \sec x = - 2 $ ,
 $ {\sec ^{ - 1}}( - 2) = x $ and the range is $ [0,\pi \} $
Now, we know that secant function is negative as
 $ \sec \left( {\dfrac{\pi }{3}} \right) = 2 $ ,
We have $ \sec \left( {\pi - \dfrac{\pi }{3}} \right) = \sec \left( {\dfrac{{2\pi }}{3}} \right) = - 2 $
Hence $ {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3} $
So, the correct answer is “ $ {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3} $ ”.

Note: To find answers to such questions use reference triangles, draw your triangle on the set of axes in the proper quadrant. Remember negative angles go in quadrant IV.
For the other answer to the sine equation, reflect the reference angle over the y-axis. A short cut method is to subtract your answer from $ \pi $ .
For the other answer to a cosine equation, reflect the reference angle over the x-axis. A short cut method is to subtract your answer from $ 2\pi $ .
For tangent, since the period given is $ \pi $ , therefore, add $ \pi $ to your first answer.