Question

If \[ \tan^{-1}\left(\frac{1}{2\sqrt2}\right) - \cos^{-1}\left(\frac{1}{\sqrt3}\right) + \sin^{-1}(x) = 0, \] then \(x=\)

• A. \(\dfrac{1}{\sqrt3}\)
• B. \(\dfrac{2}{3\sqrt3}\)
• C. \(\dfrac{1}{2\sqrt3}\)
• D. \(\dfrac{\sqrt2}{\sqrt3}\)

Hint:

For inverse trigonometric equations, assign angles to the inverse functions and determine their sine, cosine, or tangent values using right triangles. Then apply standard trigonometric identities.

Answer 1

The Correct Option is A

Concept: Use the given equation to express \(\sin^{-1}(x)\) in terms of known inverse trigonometric values and then evaluate the resulting angle.

Step 1: Rearrange the given equation. Given, \[ \tan^{-1}\left(\frac{1}{2\sqrt2}\right) - \cos^{-1}\left(\frac{1}{\sqrt3}\right) + \sin^{-1}(x) = 0. \] Therefore, \[ \sin^{-1}(x) = \cos^{-1}\left(\frac{1}{\sqrt3}\right) - \tan^{-1}\left(\frac{1}{2\sqrt2}\right). \]

Step 2: Let the angles be \(\alpha\) and \(\beta\). Let \[ \alpha=\cos^{-1}\left(\frac{1}{\sqrt3}\right). \] Then \[ \cos\alpha=\frac1{\sqrt3}. \] Hence, \[ \sin\alpha = \sqrt{1-\frac13} = \sqrt{\frac23} = \frac{\sqrt2}{\sqrt3}. \] Also, let \[ \beta=\tan^{-1}\left(\frac{1}{2\sqrt2}\right). \] Then \[ \tan\beta=\frac{1}{2\sqrt2}. \] Using a right triangle, \[ \sin\beta=\frac13, \qquad \cos\beta=\frac{2\sqrt2}{3}. \]

Step 3: Find \(\sin(\alpha-\beta)\). Since \[ \sin^{-1}(x)=\alpha-\beta, \] we have \[ x=\sin(\alpha-\beta). \] Using \[ \sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta, \] \[ x = \left(\frac{\sqrt2}{\sqrt3}\right) \left(\frac{2\sqrt2}{3}\right) - \left(\frac1{\sqrt3}\right) \left(\frac13\right). \] \[ = \frac{4}{3\sqrt3} - \frac{1}{3\sqrt3}. \] \[ = \frac{3}{3\sqrt3} = \frac1{\sqrt3}. \] Therefore, \[ \boxed{x=\frac1{\sqrt3}} \] \[ \boxed{\text{Answer = (A)}} \]