Question

If \(y = cos^2[cot^{-1}(\sqrt{\frac{1-x}{1+x}})]\) then \(\frac{dy}{dx} =\) ........

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

Hint:

Substitute x = cos 2 theta to simplify the inner expression.

Answer 1

The Correct Option is A

Step 1: Approach
Use the half-angle form directly: $\cot^{-1}u=\tan^{-1}(1/u)$.

Step 2: Double angle
Let $\varphi=\cot^{-1}u$ with $u^2=\dfrac{1-x}{1+x}$. Then $y=\cos^2\varphi=\dfrac{1+\cos2\varphi}{2}$ and $\cos2\varphi=\dfrac{u^2-1}{u^2+1}$.

Step 3: Compute
\[ \frac{u^2-1}{u^2+1}=\frac{(1-x)-(1+x)}{(1-x)+(1+x)}=-x \]
So $y=\dfrac{1-x}{2}$.

Step 4: Differentiate
$y'=-\dfrac12$, option (A).

Final Answer:
The expression simplifies to (1 - x)/2, so the derivative is -1/2, option (A). \[ \boxed{-\frac{1}{2}} \]