Question

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

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

Hint:

Use trigonometric identities to simplify inverse trigonometric expressions before differentiating. In this case: \[ \frac{1 - \cos x}{\sin x} = \tan\left(\frac{x}{2}\right) \]

Answer 1

The Correct Option is D

We are given: \[ y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \] Use the identity: \[ \frac{1 - \cos x}{\sin x} = \frac{2 \sin^2 \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} = \tan \left( \frac{x}{2} \right) \] So: \[ y = \tan^{-1} \left( \tan \left( \frac{x}{2} \right) \right) = \frac{x}{2} \quad \text{(since } -\frac{\pi}{2}<\frac{x}{2}<\frac{\pi}{2}) \] Now differentiate: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} \right) = \frac{1}{2} \] the question asks for: \[ y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \] But let’s double-check the identity: Let’s write numerator and denominator in terms of half-angle identities: \[ 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right), \quad \sin x = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \] \[ \Rightarrow \frac{1 - \cos x}{\sin x} = \frac{2 \sin^2 \left( \frac{x}{2} \right)}{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)} = \tan \left( \frac{x}{2} \right) \] So yes, confirmed: \[ y = \tan^{-1} \left( \tan \left( \frac{x}{2} \right) \right) = \frac{x}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2} \] So the correct answer is (C) \( \frac{1}{2} \) Fixing the answer:
Correct Answer Correct Answer: (C) \( \frac{1}{2} \)