Question:
If \( \theta = \cot^{-1}\sqrt{\frac{1-x}{1+x}} \), then \( \sec^2 \theta \) is equal to
If \( \theta = \cot^{-1}\sqrt{\frac{1-x}{1+x}} \), then \( \sec^2 \theta \) is equal to
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Convert inverse trig expressions into basic trig ratios before applying identities.
Updated On: Apr 21, 2026
- \( \frac{1+x}{2} \)
- \( \frac{1-x}{2} \)
- \( \frac{2}{1-x} \)
- \( x \)
- \( 2x \)
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The Correct Option is C
Solution and Explanation
Concept:
\[
\cot\theta = \sqrt{\frac{1-x}{1+x}}
\]
Step 1: Convert to \( \tan \theta \).
\[ \tan\theta = \frac{1}{\cot\theta} = \sqrt{\frac{1+x}{1-x}} \]
Step 2: Use identity.
\[ \sec^2\theta = 1 + \tan^2\theta \] \[ = 1 + \frac{1+x}{1-x} \]
Step 3: Simplify.
\[ = \frac{(1-x) + (1+x)}{1-x} = \frac{2}{1-x} \]
Step 1: Convert to \( \tan \theta \).
\[ \tan\theta = \frac{1}{\cot\theta} = \sqrt{\frac{1+x}{1-x}} \]
Step 2: Use identity.
\[ \sec^2\theta = 1 + \tan^2\theta \] \[ = 1 + \frac{1+x}{1-x} \]
Step 3: Simplify.
\[ = \frac{(1-x) + (1+x)}{1-x} = \frac{2}{1-x} \]
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