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KCET · Maths · Inverse Trigonometric Functions

If \(\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{2}\), then \(x^2\) is equal to

  1. A \(1 - y^2\)
  2. B \(\sqrt{1 - y^2}\)
  3. C \(-\sqrt{1 - y^2}\)
  4. D \(1 + y^2\)
Verified Solution

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: \(1 - y^2\)