KCET · Maths · Inverse Trigonometric Functions
If \(\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{2}\), then \(x^2\) is equal to
- A \(1 - y^2\)
- B \(\sqrt{1 - y^2}\)
- C \(-\sqrt{1 - y^2}\)
- D \(1 + y^2\)
Answer & Solution
Correct Answer
(A) \(1 - y^2\)
Step-by-step Solution
Detailed explanation
Given \(\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{2}\)
\(\sin^{-1} x = \dfrac{\pi}{2} - \sin^{-1} y\)
Using the inverse trigonometric identity \(\sin^{-1} y + \cos^{-1} y = \dfrac{\pi}{2}\), we get:
\(\sin^{-1} x = \cos^{-1} y\)
Taking sine on both sides:
\(x = \sin(\cos^{-1} y)\)
Since \(\sin(\cos^{-1} y) = \sqrt{1 - y^2}\), we have:
\(x = \sqrt{1 - y^2}\)
Squaring both sides:
\(x^2 = 1 - y^2\)
Answer: \(1 - y^2\)
\(\sin^{-1} x = \dfrac{\pi}{2} - \sin^{-1} y\)
Using the inverse trigonometric identity \(\sin^{-1} y + \cos^{-1} y = \dfrac{\pi}{2}\), we get:
\(\sin^{-1} x = \cos^{-1} y\)
Taking sine on both sides:
\(x = \sin(\cos^{-1} y)\)
Since \(\sin(\cos^{-1} y) = \sqrt{1 - y^2}\), we have:
\(x = \sqrt{1 - y^2}\)
Squaring both sides:
\(x^2 = 1 - y^2\)
Answer: \(1 - y^2\)
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