Question:
The value of \( \tan\left[ \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \right] \) is
The value of \( \tan\left[ \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \right] \) is
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Always use principal value range for inverse trigonometric functions.
Updated On: Apr 30, 2026
- \( -1 \)
- \( 0 \)
- \( 1 \)
- Infinity
- \( 2 \)
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The Correct Option is A
Solution and Explanation
Concept:
If \( \theta = \sin^{-1}(x) \), then \( \sin \theta = x \) and \( \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Step 1: Let the angle be \( \theta \). \[ \theta = \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \Rightarrow \sin \theta = -\frac{1}{\sqrt{2}} \]
Step 2: Find \( \theta \). We know: \[ \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \] Since principal value lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \[ \theta = -\frac{\pi}{4} \]
Step 3: Find tangent. \[ \tan \theta = \tan\left(-\frac{\pi}{4}\right) = -1 \]
Step 1: Let the angle be \( \theta \). \[ \theta = \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \Rightarrow \sin \theta = -\frac{1}{\sqrt{2}} \]
Step 2: Find \( \theta \). We know: \[ \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \] Since principal value lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \[ \theta = -\frac{\pi}{4} \]
Step 3: Find tangent. \[ \tan \theta = \tan\left(-\frac{\pi}{4}\right) = -1 \]
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