Question

If $ x = {\sin ^{ - 1}}K,y = {\cos ^{ - 1}}K, - 1 \leqslant k \leqslant 1, $ then the correct relationship is:-
 $
  A.x + y = 2 \\
  B.x - y = 2 \\
  C.x + y = \dfrac{\pi }{2} \\
  D.x - y = \dfrac{\pi }{2} \\
 $

Answer 1

Hint: The above given functions are inverse trigonometric functions. We recall from algebra that bijective mappings only have inverses. Since the trigonometric functions are not one-one in their natural domains, one may wonder whether these functions have inverses at all.

Complete step-by-step answer:
We can restrict the domains of the definition of these functions to appropriate subsets of their respective natural domains, then they become bijective. Therefore the above question has also restricted the domain.
The sine function restricted to $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ is one-one mapping of $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ onto $ \left[ { - 1,1} \right] $ . The inverse function $ {\sin ^{ - 1}}\left[ { - 1,1} \right] \to \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ defined by $ {\sin ^{ - 1}}K = \theta \Leftrightarrow K = \sin \theta $ for every $ K \in \left[ { - 1,1} \right] $ , is called the inverse sine function. This is also called arcsin.
Similarly inverse cosine functions like other trigonometric inverse functions are defined under following restrictions of domain as $ \left[ { - 1,1} \right] $ and its range is restricted as $ \left[ {0,\pi } \right] $ .
Now according to the question we have been given as $ x = {\sin ^{ - 1}}K,y = {\cos ^{ - 1}}k, - 1 \leqslant k \leqslant 1, $ from the above discussions we can conclude the given condition can be written as
 $
\Rightarrow \sin x = K \\
\Rightarrow \cos y = K \\
 $
Where $ x \in \left[ { - 1,1} \right] $ and $ y \in \left[ { - 1,1} \right] $ . Further the value of x+y can be calculated as,
Using the derived results of inverse trigonometric functions
 $ \therefore {\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2} $
Therefore the value is given by,
 $ x + y = {\sin ^{ - 1}}K + {\cos ^{ - 1}}K = \dfrac{\pi }{2} $
So, the correct answer is “Option C”.

Note: For any $ n \in N $ the sine function restricted to $ \left[ {(2n - 1)\dfrac{\pi }{2},(2n + 1)\dfrac{\pi }{2}} \right] = {I_n} $ a one-one map of $ {I_n} $ onto $ \left[ { - 1,1} \right] $ and has inverse. Here we are denoting the inverse function corresponding to n=0 by inverse of sine function. The inverse trigonometric functions have different graphs that are drawn with the domain and range restriction.