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Question: If \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}\), then prove that \({\sin ^{ - 1}}x = {\co...

If sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}, then prove that sin⁡−1x=cos⁡−1y{\sin ^{ - 1}}x = {\cos ^{ - 1}}y

Explanation

Solution

Here we have given sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2} and we have to prove sin⁡−1x=cos⁡−1y{\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⁡−1θ+cos⁡−1θ=π2{\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⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2} ---- eq.1
From inverse trigonometric relations,
we know
⇒sin⁡−1θ+cos⁡−1θ=π2 ⇒sin⁡−1θ=π2−cos⁡−1θ  \Rightarrow {\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2} \\\ \Rightarrow {\sin ^{ - 1}}\theta = \dfrac{\pi }{2} - {\cos ^{ - 1}}\theta \\\
Similarly,{\text{Similarly,}}on replacing θ by y\theta {\text{ by }}y
We get
⇒sin⁡−1y=π2−cos⁡−1y ——-eq.2\Rightarrow {\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y{\text{ -------eq}}{\text{.2}}
Since we need a relation betweensin⁡−1x and cos⁡−1y{\sin ^{ - 1}}x{\text{ and }}{\cos ^{ - 1}}y . So we need to change the sin⁡−1y into cos⁡−1y{\sin ^{ - 1}}y{\text{ into }}{\cos ^{ - 1}}y.
Then, we can rewrite eq.1 as
⇒sin⁡−1x+(π2−cos⁡−1y)=π2  ⇒sin⁡−1x+π2−cos⁡−1y=π2  \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,
⇒sin⁡−1x−cos⁡−1y=0\Rightarrow {\sin ^{ - 1}}x - {\cos ^{ - 1}}y = 0
And that’s why
sin⁡−1x=cos⁡−1y{\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⁡−1θ+cos⁡−1θ=π2{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}. So to solve this type of question inverse trigonometric relations must be remembered.