Inverse Trigonometric Functions — Class 12 Maths Solution

Let ${\rm{cose}}{{\rm{c}}^{ - 1}}( - \sqrt 2 ) = x \Rightarrow - \sqrt 2 = {\rm{cosec}}\;{\rm{x}}$

As we know that the range of principal value branch of ${\rm{cose}}{{\rm{c}}^{ - 1}}$ is $\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right] - \{ 0\}$

Then, $- \sqrt 2 = - {\rm{cosec}}\left( {\frac{\pi }{4}} \right) = {\rm{cosec}}\left( { - \frac{\pi }{4}} \right),$

Where $- \frac{\pi }{4} \in \left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right] - \{ 0\}$
Hence, the principal value of ${\rm{cose}}{{\rm{c}}^{ - 1}}( - \sqrt 2 )\;is\; - \frac{\pi }{4}.$