Evaluate: cot^(−1) [(((√(1−sinx))+(√(1+sinx)))/((√(1−sinx))−(√(1+sinx))))] = ?


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Question Number 51367 by rahul 19 last updated on 26/Dec/18

$E v a l u a t e :$ $\cot^{- 1} [\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}] = ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

	$c a s e - 1 a > b$ $a = s i n \frac{x}{2} b = c o s \frac{x}{2}$ $\frac{N_{r}}{D_{r}} = \frac{a - b + a + b}{a - b - a - b} = \frac{- a}{b} = - t a n \frac{x}{2} = c o t (\frac{π}{2} + \frac{x}{2})$ $s o a n s i s \frac{π}{2} + \frac{x}{2}$ $i f b > a$ $\frac{N_{r}}{D_{r}} = \frac{b - a + b + a}{b - a - b - a} = \frac{- b}{a} = - c o t \frac{x}{2} = c o t (\frac{- x}{2})$ $a n s i s \frac{- x}{2}$ $p l s c h e c k . . . x l i e s i n w h i c h q u s d r a n t ? ? ? ?$
	Commented by peter frank last updated on 26/Dec/18

	$i g o t t h e s a m e a n s$
	Answered by ajfour last updated on 26/Dec/18

	$l e t \cot θ = \frac{\sqrt{1 - \cos 2 y} + \sqrt{1 + \cos 2 y}}{\sqrt{1 - \cos 2 y} - \sqrt{1 + \cos 2 y}}$ $w h e r e x = \frac{π}{2} - 2 y \Rightarrow y = \frac{π}{4} - \frac{x}{2}$ $\Rightarrow \cot θ = \frac{\sin y + \cos y}{\sin y - \cos y} = \frac{\cos (y - \frac{π}{4})}{\sin (y - \frac{π}{4})}$ $o r \cot θ = \cot (y - \frac{π}{4}) = \cot (- \frac{x}{2})$ $\Rightarrow o r \tan θ = \tan (- \frac{x}{2})$ $θ = \cot^{- 1} [\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}]$ $= n π - \frac{x}{2} .$
	Answered by peter frank last updated on 26/Dec/18

	$\cot^{- 1} (\frac{s i n \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2}}{\sin \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2} - \cos \frac{x}{2}})$ $\cot^{- 1} (\frac{2 \sin \frac{x}{2}}{- 2 \cos \frac{x}{2}})$ $\cot^{- 1} (- \tan \frac{x}{2})$ $\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$ $\cot^{- 1} x = \frac{π}{2} - \tan^{- 1} x$ $\cot^{- 1} (- \tan \frac{x}{2}) = \frac{π}{2} - \tan^{- 1} (- \tan \frac{x}{2})$ $\frac{π}{2} + \frac{x}{2}$
	Commented by peter frank last updated on 26/Dec/18

	$r a h u l p l e a s e c h e c k$
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