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

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\le x\le 1,-1\le y\le 0
(c) y=1−x2,∣x∣<1y=\sqrt{1-{{x}^{2}}},\left| x \right|<1
(d) None of these

Explanation

Solution

Hint: In the equation given in the question, add sin⁡−1y{{\sin }^{-1}}y on both the sides of the equation will reduce the equation to sin⁡−1x=π2+sin⁡−1y{{\sin }^{-1}}x=\dfrac{\pi }{2}+{{\sin }^{-1}}y . Now, take sin on both sides of the equation you will get x=sin⁡(π2+sin⁡−1y)x=\sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}y \right) . Simplifying this expression will give us x=cos⁡(sin⁡−1y)x=\cos \left( {{\sin }^{-1}}y \right) then use the properties of inverse and solve the final expression in x and y.

Complete step-by-step answer:
The equation given in the question is:
sin⁡−1x−sin⁡−1y=π2{{\sin }^{-1}}x-{{\sin }^{-1}}y=\dfrac{\pi }{2}
Adding sin⁡−1y{{\sin }^{-1}}y on both the sides of the above equation we get,
sin⁡−1x=π2+sin⁡−1y{{\sin }^{-1}}x=\dfrac{\pi }{2}+{{\sin }^{-1}}y
Taking sin on both the sides of the above equation we get,
sin⁡(sin⁡−1x)=sin⁡(π2+sin⁡−1y)\sin \left( {{\sin }^{-1}}x \right)=\sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}y \right)
We know that if we multiply a trigonometric function by its inverse then we get 1 so sin⁡(sin⁡−1x)=1\sin \left( {{\sin }^{-1}}x \right)=1 and also the value of sin⁡(π2+θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta . Using these relations in the above equation we get,
x=cos⁡(sin⁡−1y)x=\cos \left( {{\sin }^{-1}}y \right)
In the above equation, let us assume that sin⁡−1y=φ{{\sin }^{-1}}y=\varphi then taking sin on both the sides we get y=sin⁡φy=\sin \varphi and we need the value of cos⁡φ\cos \varphi which is equal to 1−y2\sqrt{1-{{y}^{2}}} because cos⁡φ=1−sin⁡2φ\cos \varphi =\sqrt{1-{{\sin }^{2}}\varphi } .
x=cos⁡φx=\cos \varphi
Substituting the value of cos⁡φ=1−sin⁡2φ\cos \varphi =\sqrt{1-{{\sin }^{2}}\varphi } in the above equation we get,
x=1−sin⁡2φx=\sqrt{1-{{\sin }^{2}}\varphi }
And we have shown above that the value of sin⁡φ=y\sin \varphi =y so substituting this value in the above equation.
x=1−y2x=\sqrt{1-{{y}^{2}}}
Squaring on both the sides of the above equation we get,
x2=1−y2{{x}^{2}}=1-{{y}^{2}}
None of the option given in the question matches the above form of equation so rearranging the above expression.
x2+y2=1{{x}^{2}}+{{y}^{2}}=1
Now, the above equation matches with the option (a).
Hence, the correct option is (a).

Note: In the above solution, we have shown that sin⁡(π2+θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta . This is the property of trigonometric angles that if we add odd multiples of π2\dfrac{\pi }{2} to the given angle then sin⁡\sin becomes cos⁡\cos and vice versa so here, sin⁡(π2+θ)\sin \left( \dfrac{\pi }{2}+\theta \right) becomes cos⁡θ\cos \theta and the sign of this conversion is positive because we know that sine of a given angle in the first and second quadrant is positive.