Question

If
\[ x=\cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), \quad y=\sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right), \] then \(\frac{dy}{dx}\) is equal to

• A. \(0\)
• B. \(\tan t\)
• C. \(1\)
• D. \(\sin t\cos t\)

Hint:

If two inverse trig expressions produce the same sine/cosine values, they may represent the same angle. Then \(y=x\Rightarrow dy/dx=1\).

Answer 1

The Correct Option is C

Step 1: Simplify expressions using trigonometric interpretation.
Let:
\[ \cos x=\frac{1}{\sqrt{1+t^2}} \]
Then:
\[ \sin x=\sqrt{1-\cos^2x} =\sqrt{1-\frac{1}{1+t^2}} =\sqrt{\frac{t^2}{1+t^2}} =\frac{t}{\sqrt{1+t^2}} \]
Step 2: Compare with \(y\).
\[ y=\sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) =\sin^{-1}(\sin x) \Rightarrow y=x \]
Step 3: Differentiate.
If \(y=x\), then:
\[ \frac{dy}{dx}=1 \]
Final Answer:
\[ \boxed{1} \]