Question

If $y = \tan^{-1}\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)$, then $\frac{dy}{dx} =$

• A. $1$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Hint:

Memorize standard reductions to save precious time: $\sqrt{\frac{1+\cos x}{1-\cos x}} = \cot\left(\frac{x}{2}\right)$. Since $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$, the function reduces instantly to $\frac{\pi}{2} - \frac{x}{2}$. Its derivative is simply the coefficient of $x$, which is $-\frac{1}{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to differentiate y = tan⁻¹(√((1+cos x)/(1–cos x))) by first simplifying the trigonometric expression inside the radical.

Step 2: Key Formula or Approach:
Use half-angle identities: 1+cos x = 2cos²(x/2) and 1–cos x = 2sin²(x/2), then simplify the inverse trigonometric composite.

Step 3: Detailed Explanation:
Substituting gives y = tan⁻¹(√(cot²(x/2))) = tan⁻¹(cot(x/2)) = tan⁻¹(tan(π/2 – x/2)) = π/2 – x/2. Differentiating yields dy/dx = –1/2.

Step 4: Final Answer:
The derivative is –1/2, matching option (D).