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: Formula
Use the identity $\cot^{-1} \theta + \tan^{-1} \theta = \pi/2$.

Step 2: Analysis
From identity: $\cot^{-1}(\sqrt{\cos \alpha}) = \pi/2 - \tan^{-1}(\sqrt{\cos \alpha})$.
Substitute in equation: $(\pi/2 - \tan^{-1}\sqrt{\cos \alpha}) - \tan^{-1}\sqrt{\cos \alpha} = x$
$\pi/2 - 2\tan^{-1}\sqrt{\cos \alpha} = x \implies 2\tan^{-1}\sqrt{\cos \alpha} = \pi/2 - x$.

Step 3: Calculation
Apply $\tan$ on both sides: $\tan(2\tan^{-1}\sqrt{\cos \alpha}) = \tan(\pi/2 - x) = \cot x$.
$\frac{2\sqrt{\cos \alpha}}{1 - \cos \alpha} = \cot x$.
Therefore, $\tan x = \frac{1 - \cos \alpha}{2\sqrt{\cos \alpha}}$. (Using $\tan 2\theta = 2t/(1-t^2)$).
Using $\sin x$ formula from $\tan x$: $\sin x = \frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = \tan^2(\alpha/2)$.

Step 4: Conclusion
Hence, $\sin x = \tan^2(\alpha/2)$. Final Answer: (D)