Question

If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}$, then prove that ${\sin ^{ - 1}}x = {\cos ^{ - 1}}y$

Answer 1

Hint: Here we have given ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}$ and we have to prove ${\sin ^{ - 1}}x = {\cos ^{ - 1}}y$ we can see that LHS of our proof is already in our question so we have to only get RHS using properties of inverse trigonometric functions like ${\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}$ , using this formulae we will proceed in this question.


Complete step-by-step answer:
Given
${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}$ ---- eq.1
From inverse trigonometric relations,
we know
$
   \Rightarrow {\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2} \\
   \Rightarrow {\sin ^{ - 1}}\theta = \dfrac{\pi }{2} - {\cos ^{ - 1}}\theta \\
 $
${\text{Similarly,}}$on replacing $\theta {\text{ by }}y$
We get
$ \Rightarrow {\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y{\text{ -------eq}}{\text{.2}} $
Since we need a relation between${\sin ^{ - 1}}x{\text{ and }}{\cos ^{ - 1}}y$ . So we need to change the ${\sin ^{ - 1}}y{\text{ into }}{\cos ^{ - 1}}y$.
Then, we can rewrite eq.1 as
$
   \Rightarrow {\sin ^{ - 1}}x + (\dfrac{\pi }{2} - {\cos ^{ - 1}}y) = \dfrac{\pi }{2}{\text{ }} \\
   \Rightarrow {\sin ^{ - 1}}x + \dfrac{\pi }{2} - {\cos ^{ - 1}}y = \dfrac{\pi }{2} \\
$
( using eq.2 )
Now on cancel out we get,
$ \Rightarrow {\sin ^{ - 1}}x - {\cos ^{ - 1}}y = 0$
And that’s why
${\sin ^{ - 1}}x = {\cos ^{ - 1}}y$
Hence Proved

Note: Whenever you get this type of problem the key concept of solving the problem that you have to learn inverse trigonometric relations and their applications too like in this problem we require the relation ${\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}$. So to solve this type of question inverse trigonometric relations must be remembered.