If U=cot−1⁡cos⁡2θ−tan−1⁡cos⁡2θ, then sin U equals

We have,U=cot−1⁡cos⁡2θ−tan−1⁡cos⁡2θ⇒ U=π2−tan−1⁡cos⁡2θ−tan−1⁡cos⁡2θ⇒ U=π2−2tan−1⁡cos⁡2θ⇒ U=π2−tan−1⁡2cos⁡2θ1−cos⁡2θ⇒ U=π2−tan−1⁡cos⁡2θsin2⁡θ⇒ tan⁡π2−U=cos⁡2θsin2⁡θ⇒ cot⁡U=cos⁡2θsin2⁡θ⇒ sin⁡U=11+cot2⁡θ=11+cos⁡2θsin4⁡θ⇒ sin⁡U=sin2⁡θsin4⁡θ+1−2sin2⁡θ⇒ sin⁡U=sin2⁡θ1−sin2⁡θ=tan2⁡θ

If U=cot−1⁡cos⁡2θ−tan−1⁡cos⁡2θ, then sin U equals

We have,U=cot−1⁡cos⁡2θ−tan−1⁡cos⁡2θ⇒ U=π2−tan−1⁡cos⁡2θ−tan−1⁡cos⁡2θ⇒ U=π2−2tan−1⁡cos⁡2θ⇒ U=π2−tan−1⁡2cos⁡2θ1−cos⁡2θ⇒ U=π2−tan−1⁡cos⁡2θsin2⁡θ⇒ tan⁡π2−U=cos⁡2θsin2⁡θ⇒ cot⁡U=cos⁡2θsin2⁡θ⇒ sin⁡U=11+cot2⁡θ=11+cos⁡2θsin4⁡θ⇒ sin⁡U=sin2⁡θsin4⁡θ+1−2sin2⁡θ⇒ sin⁡U=sin2⁡θ1−sin2⁡θ=tan2⁡θ

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If $U = \cot^{- 1} \sqrt{\cos 2 θ} - \tan^{- 1} \sqrt{\cos 2 θ}$ , then $\sin U$ equals

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$\sin^{2} θ$

$\cos^{2} θ$

$\tan^{2} θ$

$\tan^{2} 2 θ$

answer is C.

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Detailed Solution

We have,

$\begin{array}{l} U = \cot^{- 1} \sqrt{\cos 2 θ} - \tan^{- 1} \sqrt{\cos 2 θ} \\ \Rightarrow U = \frac{π}{2} - \tan^{- 1} \sqrt{\cos 2 θ} - \tan^{- 1} \sqrt{\cos 2 θ} \\ \Rightarrow U = \frac{π}{2} - 2 \tan^{- 1} \sqrt{\cos 2 θ} \\ \Rightarrow U = \frac{π}{2} - \tan^{- 1} (\frac{2 \sqrt{\cos 2 θ}}{1 - \cos 2 θ}) \\ \Rightarrow U = \frac{π}{2} - \tan^{- 1} (\frac{\sqrt{\cos 2 θ}}{\sin^{2} θ}) \\ \Rightarrow \tan (\frac{π}{2} - U) = \frac{\sqrt{\cos 2 θ}}{\sin^{2} θ} \\ \Rightarrow \cot U = \frac{\sqrt{\cos 2 θ}}{\sin^{2} θ} \\ \Rightarrow \sin U = \frac{1}{\sqrt{1 + \cot^{2} θ}} = \frac{1}{\sqrt{1 + \frac{\cos 2 θ}{\sin^{4} θ}}} \\ \Rightarrow \sin U = \frac{\sin^{2} θ}{\sqrt{\sin^{4} θ + 1 - 2 \sin^{2} θ}} \\ \Rightarrow \sin U = \frac{\sin^{2} θ}{|1 - \sin^{2} θ|} = \tan^{2} θ \end{array}$