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

If sin⁡−1x−sin⁡−1y=π2{\sin ^{ - 1}}x - {\sin ^{ - 1}}y = \dfrac{\pi }{2}, then
A) x2+y2=1{x^2} + {y^2} = 1
B) y=−1−x2,0⩽x⩽1,−1⩽y⩽0y = - \sqrt {1 - {x^2}} ,0 \leqslant x \leqslant 1, - 1 \leqslant y \leqslant 0
C) y=1−x2,∣x∣<1y = \sqrt {1 - {x^2}} ,\left| x \right| < 1
D) None of these

Explanation

Solution

According to given in the question we have to determine the value of sin⁡−1x−sin⁡−1y=π2{\sin ^{ - 1}}x - {\sin ^{ - 1}}y = \dfrac{\pi }{2} so, first of all we have to take the term sin⁡−1x{\sin ^{ - 1}}x to the right hand side and after then we have to take sin in the both sides of the trigonometric expression.
Now, to solve the trigonometric expression we have to use the formula as mentioned below:

Formula used: ⇒sin⁡−1(sin⁡x)=x..............(A) ⇒cos⁡−1(cos⁡x)=x.............(B) \Rightarrow {\sin ^{ - 1}}(\sin x) = x..............(A) \\\ \Rightarrow {\cos ^{ - 1}}(\cos x) = x.............(B)
Now, to solve the obtained expression we have to convert sin⁡−1{\sin ^{ - 1}} in form of cos⁡−1{\cos ^{ - 1}} which can be converted with the help of the formula as mentioned below:
⇒sin⁡−1x=cos⁡−11−x2................(C)\Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} ................(C)
Now, on solving the obtained algebraic expression we can obtain the correct option.

Complete step-by-step solution:
Step 1: First of all we have to take the term sin⁡−1x{\sin ^{ - 1}}x to the right hand side and after then we have to take sin on both sides of the trigonometric expression as mentioned in the solution hint. Hence,
⇒sin⁡−1y=−(π2−sin⁡−1x)...................(1)\Rightarrow {\sin ^{ - 1}}y = - \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right)...................(1)
Step 2: Now, we have to take sin on both sides of the expression (1) as obtained in the solution step 1 and as mentioned in the solution hint.
⇒sin⁡sin⁡−1y=−sin⁡(π2−sin⁡−1x) ⇒sin⁡sin⁡−1y=−cos⁡(sin⁡−1x)..................(2) \Rightarrow \sin {\sin ^{ - 1}}y = - \sin \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right) \\\ \Rightarrow \sin {\sin ^{ - 1}}y = - \cos ({\sin ^{ - 1}}x)..................(2)
Step 3: Now, to solve the trigonometric expression (2) as obtained in the step 2 we have to use the formula (A) as mentioned in the solution hint. Hence,
⇒y=−cos⁡(sin⁡−1x)...................(3)\Rightarrow y = - \cos \left( {{{\sin }^{ - 1}}x} \right)...................(3)
Step 4: Now, to solve the trigonometric expression (3) as obtained in the solution step 3 we have to use the formula (C) as mentioned in the solution hint.
⇒y=−cos⁡cos⁡−11−x2\Rightarrow y = - \cos {\cos ^{ - 1}}\sqrt {1 - {x^2}}…………………….(4)
Step 5: Now, to solve the trigonometric expression (4) as obtained in the solution step 4 we have to use the formula (B) as mentioned in the solution hint.
⇒y=−1−x2\Rightarrow y = - \sqrt {1 - {x^2}}
⇒−1⩽y⩽0\Rightarrow - 1 \leqslant y \leqslant 0
Final solution: Hence, with the help of formula (A), (B), and (C) we have obtained the value of the given trigonometric expression which is −1⩽y⩽0 - 1 \leqslant y \leqslant 0.

Therefore option (B) is correct.

Note: To solve the given trigonometric expression it is not necessary that we have to take sin⁡−1x{\sin ^{ - 1}}x to the right hand side of the expression we can also take sin⁡−1y{\sin ^{ - 1}}y to the right hand side of the expression.
If we take sin⁡−1x{\sin ^{ - 1}}x in the right hand side of the expression we will get the answer in form of y or if we take sin⁡−1y{\sin ^{ - 1}}y to the right hand side and on solving we will obtain the answer in form of x.