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# If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2},$ then ${x^2}$ is equal to(a)$1 - {y^2}$(b)${y^2}$(c)$0$(d)$\sqrt {1 - y}$

Last updated date: 25th Jun 2024
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Hint:As we know that the above given question is related to trigonometric expression, sine and cosine are trigonometric ratios. Here we have to find the value of ${x^2}$, so first of all we have to take the term ${\sin ^{ - 1}}y$ to the right hand side and then after that we have to multiply with sin in the both sides of the trigonometric expression. We can convert the equation into basic trigonometric equations by applying the trigonometric identities.
As per the given question we have to solve the equation ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{\pi }{2}$ .
We will first take the term ${\sin ^{ - 1}}y$ to the right hand side and we get: ${\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}y$. After this we will multiply with the term sin in both the left hand side and the right hand side of the equation, $\sin ({\sin ^{ - 1}}x) = \sin \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}y} \right)$. We will use this trigonometric formula to solve it further, ${\sin ^{ - 1}}(\sin x) = x$ and another identity is $\sin (90 - \theta ) = \cos \theta$, now applying this we get :
$x = \cos ({\sin ^{ - 1}}y)$, as we know that $(\dfrac{\pi }{2} - \theta )$ can be expressed as $(90 - \theta )$.
We can use the another trigonometric formula i.e. ${\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}}$, this can be done by converting the term sin into cosine, so we get: $x = \cos \left( {{{\cos }^{ - 1}}\sqrt {1 - {y^2}} } \right) = \sqrt {1 - {y^2}}$.
So by using the basic rule of trigonometry and squaring both the sides we get, ${x^2} = 1 - {y^2}$.
Hence the correct option is (a)$1 - {y^2}$.
Note: Before solving this kind of question we should have the proper knowledge of all trigonometric ratios, identities and their formulas. To solve this trigonometric expression it is not necessary to take only ${\sin ^{ - 1}}y$ to the right hand side, we can also take another left hand term to the right hand side of the expression.