Question:
If \( f(x) = \sin^{-1} \left( \frac{1 - \cos 2x}{2 \sin x} \right) \), then \( |f'(x)| \) is equal to
If \( f(x) = \sin^{-1} \left( \frac{1 - \cos 2x}{2 \sin x} \right) \), then \( |f'(x)| \) is equal to
Show Hint
Always simplify trigonometric expressions before differentiating inverse functions.
Updated On: Jul 5, 2026
- \( |\sin x| \)
- \( x \)
- \( 0 \)
- \( |\cos x| \)
- \( 1 \)
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The Correct Option is
Solution and Explanation
Concept:
Use trigonometric identity:
\[
1-\cos 2x = 2\sin^2 x
\]
Step 1: Simplify inside inverse sine
\[ \frac{1-\cos 2x}{2\sin x} = \frac{2\sin^2 x}{2\sin x} = \sin x \]
Step 2: Reduce function
\[ f(x) = \sin^{-1}(\sin x) \]
Step 3: Understand behaviour
For principal values: \[ \sin^{-1}(\sin x) = x \text{ (in valid domain)} \] Thus derivative: \[ f'(x)=1 \text{ or } -1 \]
Step 4: Take modulus
\[ |f'(x)| = 1 \]
Step 5: Final Answer
\[ \boxed{1} \]
Step 1: Simplify inside inverse sine
\[ \frac{1-\cos 2x}{2\sin x} = \frac{2\sin^2 x}{2\sin x} = \sin x \]
Step 2: Reduce function
\[ f(x) = \sin^{-1}(\sin x) \]
Step 3: Understand behaviour
For principal values: \[ \sin^{-1}(\sin x) = x \text{ (in valid domain)} \] Thus derivative: \[ f'(x)=1 \text{ or } -1 \]
Step 4: Take modulus
\[ |f'(x)| = 1 \]
Step 5: Final Answer
\[ \boxed{1} \]
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