Question

If $ {{\sin }^{-1}}x-{{\sin }^{-1}}y=\dfrac{\pi }{2} $ , then
(a) $ {{x}^{2}}+{{y}^{2}}=1 $
(b) $ y=-\sqrt{1-{{x}^{2}}},0\le x\le 1,-1\le y\le 0 $
(c) $ y=\sqrt{1-{{x}^{2}}},\left| x \right|<1 $
(d) None of these

Answer 1

Hint: In the equation given in the question, add $ {{\sin }^{-1}}y $ on both the sides of the equation will reduce the equation to $ {{\sin }^{-1}}x=\dfrac{\pi }{2}+{{\sin }^{-1}}y $ . Now, take sin on both sides of the equation you will get $ x=\sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}y \right) $ . Simplifying this expression will give us $ 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 }^{-1}}x-{{\sin }^{-1}}y=\dfrac{\pi }{2} $
Adding $ {{\sin }^{-1}}y $ on both the sides of the above equation we get,
 $ {{\sin }^{-1}}x=\dfrac{\pi }{2}+{{\sin }^{-1}}y $
Taking sin on both the sides of the above equation we get,
 $ \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 \left( {{\sin }^{-1}}x \right)=1 $ and also the value of $ \sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta $ . Using these relations in the above equation we get,
 $ x=\cos \left( {{\sin }^{-1}}y \right) $
In the above equation, let us assume that $ {{\sin }^{-1}}y=\varphi $ then taking sin on both the sides we get $ y=\sin \varphi $ and we need the value of $ \cos \varphi $ which is equal to $ \sqrt{1-{{y}^{2}}} $ because $ \cos \varphi =\sqrt{1-{{\sin }^{2}}\varphi } $ .
 $ x=\cos \varphi $
Substituting the value of $ \cos \varphi =\sqrt{1-{{\sin }^{2}}\varphi } $ in the above equation we get,
 $ x=\sqrt{1-{{\sin }^{2}}\varphi } $
And we have shown above that the value of $ \sin \varphi =y $ so substituting this value in the above equation.
 $ x=\sqrt{1-{{y}^{2}}} $
Squaring on both the sides of the above equation we get,
 $ {{x}^{2}}=1-{{y}^{2}} $
None of the option given in the question matches the above form of equation so rearranging the above expression.
 $ {{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 \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta $ . This is the property of trigonometric angles that if we add odd multiples of $ \dfrac{\pi }{2} $ to the given angle then $ \sin $ becomes $ \cos $ and vice versa so here, $ \sin \left( \dfrac{\pi }{2}+\theta \right) $ becomes $ \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.