KCET · Maths · Inverse Trigonometric Functions
If \( \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \), then \( x^{2} \) is equal to
- A \( 1-y^{2} \)
- B \( y^{2} \)
- C \( 00 \)
- D \( \sqrt{1-y} \)
Answer & Solution
Correct Answer
(A) \( 1-y^{2} \)
Step-by-step Solution
Detailed explanation
Given that,
\( \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \rightarrow(1) \)
We know that,
\( \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \rightarrow(2) \)
From Eqs. (1) and (2), we get
\( \sin ^{-1} y=\cos ^{-1} x \) \( x=\cos \left(\sin ^{-1} y\right) \)
We know that, \( \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}} \)
So, \( \cos \left(\cos ^{-1} \sqrt{1-y^{2}}\right)=\sqrt{1-y^{2}} \)
\( \Rightarrow x^{2}=\left(1-y^{2}\right) \)
\( \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \rightarrow(1) \)
We know that,
\( \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \rightarrow(2) \)
From Eqs. (1) and (2), we get
\( \sin ^{-1} y=\cos ^{-1} x \) \( x=\cos \left(\sin ^{-1} y\right) \)
We know that, \( \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}} \)
So, \( \cos \left(\cos ^{-1} \sqrt{1-y^{2}}\right)=\sqrt{1-y^{2}} \)
\( \Rightarrow x^{2}=\left(1-y^{2}\right) \)
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