Question

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

• A. \(\sqrt{\frac{1-x^2}{1-2x^2}}\)
• B. \(\frac{x}{\sqrt{1-2x^2}}\)
• C. \(\sqrt{\frac{1-2x^2}{1-x^2}}\)
• D. \(\frac{\sqrt{1-2x^2}}{x}\) \bigskip

Hint:

Convert inverse trig into triangle ratios instead of direct identities.

Answer 1

The Correct Option is C

Concept: Convert inverse trig expressions step by step using triangles.

Step 1: Let \(\theta=\sin^{-1}x\).
\[ \sin\theta=x,\quad \cos\theta=\sqrt{1-x^2} \] \[ \tan(\sin^{-1}x)=\frac{x}{\sqrt{1-x^2}} \]

Step 2: Let \(\phi=\cos^{-1}(\tan\theta)\).
So \[ \cos\phi=\frac{x}{\sqrt{1-x^2}} \]

Step 3: Compute cotangent.
\[ \cot\phi=\frac{\cos\phi}{\sqrt{1-\cos^2\phi}} \] \[ =\sqrt{\frac{1-2x^2}{1-x^2}} \] \[ \boxed{(C)} \]