VerifiedHint: Whenever you come up with this type of problem then first convert inverse trigonometry function into normal trigonometry function, and then compare. Remember the range of principal branches of all inverse trigonometric functions.
Complete step by step answer:
As, we know that the Principal Value of Inverse Trigonometric Functions at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch.
As we know that here we have to find principal value of $\cos ec^{ - 1}(x)$, where x = $- \sqrt 2$
$\Rightarrow$ Let $\cos ec^{ - 1}( - \sqrt 2 ) = y$ - (Eq 1)
So, taking $\cos ec$ both sides of the equation 1 we get,
$\Rightarrow \cos ec(y) = - \sqrt 2$ - (Eq 2)
And as we know that $\cos ec\left( {\dfrac{\pi }{4}} \right) = \sqrt 2$.
So, $\sqrt 2$ in equation 2 can be written as $\cos ec\left( {\dfrac{\pi }{4}} \right)$
So, equation 2 becomes $\cos ec(y) = - \cos ec\left( {\dfrac{\pi }{4}} \right)$
$\Rightarrow \cos ec(y) = - \cos ec\left( {\dfrac{\pi }{4}} \right)$ -(Eq 3)
And as we know that $\cos ec\left( { - \theta } \right) = - \cos ec\left( \theta \right)$
So, equation 3 becomes,
$\Rightarrow \cos ec(y) = \cos ec\left( { - \dfrac{\pi }{4}} \right)$
And as we know range of principal branch of $\cos e{c^{ - 1}}\left( \theta \right){\text{ }}is{\text{ }}\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] - \{ 0\}$
$\Rightarrow$ and $cosec\left( { - \dfrac{\pi }{4}} \right) = - \sqrt 2$.
$\Rightarrow$ So, therefore principle value of $\cos e{c^{ - 1}}( - \sqrt 2 ){\text{ }}i{\text{s }} - \dfrac{\pi }{4}$
NOTE: - There are two different ways to find the inverse of a function. One is graphing and the other is algebra. However, when finding the inverse of trigonometric functions, it is easy to find the inverse of a trigonometric function through graphing.
