Question

If \(y=\sin ^2\left(\cot ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)

• A. -1
• B. 1
• C. \(-\frac{1}{4}\)
• D. \(\frac{1}{2}\)

Hint:

Simplify sin squared of cot inverse t using 1 over 1 plus t squared before differentiating.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
If \(\theta=\cot^{-1}t\), then \(\cot\theta=t\) and \(\sin^2\theta=\dfrac{1}{1+\cot^2\theta}=\dfrac{1}{1+t^2}\).

Step 2: Simplify y
Here \(t=\sqrt{\dfrac{1-x}{1+x}}\), so \(t^2=\dfrac{1-x}{1+x}\).
\(y=\dfrac{1}{1+\frac{1-x}{1+x}}=\dfrac{1+x}{2}\)

Step 3: Differentiate
\(\dfrac{dy}{dx}=\dfrac{1}{2}\)

Step 4: Compare the options
The value \(\dfrac{1}{2}\) matches the last option.

Final Answer:
The derivative is one half. \[ \boxed{\text{(D) }\dfrac{1}{2}} \]