If u=cot−1⁡ cos⁡ α−tan−1⁡ cos⁡ α, then sin u is equal to

Let cos⁡ α=tan⁡ y. Then tan−1⁡ cos⁡ α=yand cot−1⁡ cos⁡ α=π/2−y. Therefore u=(π/2)−y−y=(π/2)−2y.⇒sin⁡ u=cos⁡ 2y=1−tan2⁡ y1+tan2⁡ y=1−cos⁡ α1+cos⁡ α=2sin2⁡ (α/2)2cos2⁡ (α/2)=tan2⁡α2

If u=cot−1⁡ cos⁡ α−tan−1⁡ cos⁡ α, then sin u is equal to

Let cos⁡ α=tan⁡ y. Then tan−1⁡ cos⁡ α=yand cot−1⁡ cos⁡ α=π/2−y. Therefore u=(π/2)−y−y=(π/2)−2y.⇒sin⁡ u=cos⁡ 2y=1−tan2⁡ y1+tan2⁡ y=1−cos⁡ α1+cos⁡ α=2sin2⁡ (α/2)2cos2⁡ (α/2)=tan2⁡α2

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If $u = \cot^{- 1} \sqrt{\cos α} - \tan^{- 1} \sqrt{\cos α}$ , then sin u is equal to

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$\cot^{2} (α / 2)$

$\tan^{2} (α / 2)$

$\cot^{2} α$

$\tan^{2} α$

answer is A.

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Detailed Solution

Let $\sqrt{\cos α} = \tan y . Then \tan^{- 1} \sqrt{\cos α} = y$

and $\cot^{- 1} \sqrt{\cos α} = π / 2 - y$ . Therefore $u = (π / 2) - y -$ $y = (π / 2) - 2 y$ .

$\begin{array}{l} \Rightarrow \sin u = \cos 2 y \\ = \frac{1 - \tan^{2} y}{1 + \tan^{2} y} = \frac{1 - \cos α}{1 + \cos α} = \frac{2 \sin^{2} (α / 2)}{2 \cos^{2} (α / 2)} = \tan^{2} \frac{α}{2} \end{array}$