Question

If $\cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$, then the value of $\sin x$ is}

• A. $\cot^{2}\frac{\alpha}{2}$
• B. $\cot \alpha$
• C. $\tan\frac{\alpha}{2}$
• D. $\tan^{2}\frac{\alpha}{2}$

Hint:

$\sin x = \tan^2(\alpha/2)$ is a direct result of the half-angle formula for $\cos \alpha$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for $\sin x$ given an equation involving inverse trigonometric functions of $\sqrt{\cos \alpha}$.

Step 2: Key Formula or Approach:
1. $\cot^{-1} u + \tan^{-1} u = \frac{\pi}{2}$.
2. $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$.
3. Half-angle formula: $\tan^2 \frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos \alpha}$.

Step 3: Detailed Explanation:
Let $\theta = \tan^{-1}(\sqrt{\cos \alpha})$. Then $\tan \theta = \sqrt{\cos \alpha}$.
From $\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u$, the equation becomes: \[ (\frac{\pi}{2} - \theta) - \theta = x \] \[ x = \frac{\pi}{2} - 2\theta \] Taking sine on both sides: \[ \sin x = \sin(\frac{\pi}{2} - 2\theta) = \cos 2\theta \] Using the formula for $\cos 2\theta$: \[ \sin x = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \] Since $\tan \theta = \sqrt{\cos \alpha}$, $\tan^2 \theta = \cos \alpha$.
\[ \sin x = \frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \tan^2 \frac{\alpha}{2} \]

Step 4: Final Answer:
The value of $\sin x$ is $\tan^2 \frac{\alpha}{2}$.