what is the value of dy/dx if y = cot^(−1) [(((√(1+sinx))+(√(1−sinx)))/( (√(1+sinx))−(√(1−sinx))))]; x∈(0,(π/2)) please give complete explanation that why can we not take [sinx/2−cosx/2]^2 = 1−sinx


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Question Number 192758 by vishal1234 last updated on 26/May/23

$w h a t i s t h e v a l u e o f d y / d x i f$ $y = {c o t}^{- 1} [\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}]; x \in (0, \frac{π}{2})$ $p l e a s e g i v e c o m p l e t e e x p l a n a t i o n t h a t$ $w h y c a n w e n o t t a k e$ ${[s i n x / 2 - c o s x / 2]}^{2} = 1 - s i n x$
	Answered by a.lgnaoui last updated on 26/May/23
	$for the second question (sin (x/2)−cos (x/2) )^2 =(sin (x/2))^2 +(cos (x/2))^2 −2(sin (x/2))(cos (x/2)) = 1±2[(sin (x/2))(√(1−(sin (x/2))^2 )) posons u=sin (x/2) z=(sin (x/2)−cos (x/2))^2 { ((z=1+2u(√(1−u^2 )) (1))),((z=1−2u(√(1−u^2 )) (2))) :} 1+2u(√(1−u^2 )) =1+2sin (x/2)cos (x/2) =1+sin x 1−2u(√(1−u^2 )) =1−sin x ⇒(sin (x/2)−cos (x/2))^2 =1+∣sin x∣ •−(π/2)+2kπ<x<+(π/2)+2kπ sin x>0 z=1+sin x •(π/2)+2kπ<x<((3π)/2)+2kπ sin x<0 z=1−sin x donc z n est pas toujours egale ( 1−sinx)$
	$for the second question$ ${(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2} = {(\sin \frac{x}{2})}^{2} + {(\cos \frac{x}{2})}^{2}$ $- 2 (\sin \frac{x}{2}) (\cos \frac{x}{2})$ $= 1 \pm 2 [(\sin \frac{x}{2}) \sqrt{1 - {(\sin \frac{x}{2})}^{2}}$ $posons u = \sin \frac{x}{2} z = {(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2}$ ${\begin{cases} z = 1 + 2 u \sqrt{1 - u^{2}} (1) \\ z = 1 - 2 u \sqrt{1 - u^{2}} (2) \end{cases}$ $1 + 2 u \sqrt{1 - u^{2}} = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ $= 1 + \sin x$ $1 - 2 u \sqrt{1 - u^{2}} = 1 - \sin x$ $\Rightarrow {(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2} = 1 + ∣ \sin x ∣$ $∙ - \frac{π}{2} + 2 k π < x < + \frac{π}{2} + 2 k π$ $\sin x > 0 z = 1 + \sin x$ $∙ \frac{π}{2} + 2 k π < x < \frac{3 π}{2} + 2 k π$ $\sin x < 0 z = 1 - \sin x$ $donc z n est pas toujours egale (1 - sinx)$
	Commented by a.lgnaoui last updated on 26/May/23

	$- \frac{π}{4} < \frac{x}{2} < \frac{π}{4} cas 1$ $\frac{π}{4} < \frac{x}{2} < π + \frac{π}{4} cas 2$ $∙ \sin \frac{x}{2} \cos \frac{x}{2} > 0 cas 1 \Rightarrow z = 1 - \sin x$ $∙ \sin \frac{x}{2} \cos \frac{x}{2} < 0 cas 2 \Rightarrow z = 1 + \sin x$
	Answered by mahdipoor last updated on 26/May/23

	$w h y c a n n o t t a k e ?!$ $t a k e t h a t, b o t p a y a t t e n t i o n :$ $\sqrt{1 - s i n x} = \sqrt{{(s i n \frac{x}{2} - c o s \frac{x}{2})}^{2}} =$ $∣ s i n \frac{x}{2} - c o s \frac{x}{2} ∣= c o s \frac{x}{2} - s i n \frac{x}{2} a t 0 < x < \frac{π}{2}$ $a n d$ $\sqrt{1 + s i n x} = c o s \frac{x}{2} + s i n \frac{x}{2}$ $y = {c o t}^{- 1} (\frac{c o s \frac{x}{2}}{s i n \frac{x}{2}}) = {c o t}^{- 1} (c o t (\frac{x}{2})) = \frac{x}{2} \Rightarrow \frac{d y}{d x} = \frac{1}{2}$
	Answered by a.lgnaoui last updated on 26/May/23

	$First question$ $\cot y = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}$ $= \frac{{[\sqrt{1 + \sin x} + \sqrt{1 - \sin x}]}^{2}}{2 \sin x}$ $= \frac{1 + ∣ \cos x ∣}{\sin x}$ $0 < x < \frac{π}{2} \Rightarrow \cos x > 0$ $\Rightarrow \cot y = \frac{1 + \cos x}{\sin x}$ $\frac{d (\cot y)}{dx} = \frac{d (\cot y)}{dy} \times \frac{dy}{dx}$ $- \frac{\cos x + 1}{\sin^{2} x} = - \frac{1}{\sin^{2} y} \times \frac{dy}{dx}$ $\frac{1 + \cos x}{\sin^{2} x} \sin^{2} y = \frac{dy}{dx}$ $\frac{dy}{dx} = [\frac{1 + \cos x}{\sin^{2} x}] (1 + \cot^{2} y)$ $= \frac{1 + \cos x}{\sin^{2} x} (1 + \frac{1 + \cos x}{\sin x})$ $= \frac{1 + \cos x}{\sin^{2} x} + \frac{{(1 + \cos x)}^{2}}{\sin^{3} x}$ $\frac{d y}{dx} = \frac{(1 + \cos x) (\sin x + \cos x + 1)}{\sin^{3} x}$
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