First slide

Properties of ITF

Question

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

Moderate

A

$\sin^{2} θ$
B

$\cos^{2} θ$
C

$\tan^{2} θ$
D

$\tan^{2} 2 θ$

Solution

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

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