Answer
384.6k+ views
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.
Complete step by step solution:
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.
Complete step by step solution:
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)