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Question Number 40662 by Tinkutara last updated on 25/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

$p l s c h e c k t h e q u e s t i o n . . . i t s e e m s w r o n g . . .$
Commented by maxmathsup by imad last updated on 25/Jul/18
$if y_1 means derivative let w(x)=((√(a^2 −b^2 ))/2)y(x)⇒w(x)=arctan(((a−b)/(a+b))tan((x/2))) ⇒w^′ (x)=((((a−b)/(2(a+b))){1+tan^2 ((x/2))})/(1+(((a−b)^2 )/((a+b)^2 ))tan^2 ((x/2)))) =(((a−b))/(2(a+b)cos^2 ((x/2)){(((a+b)^2 +(a−b)^2 tan^2 ((x/2)))/((a+b)^2 ))})) =((a^2 −b^2 )/(2cos^2 ((x/2)){(a+b)^2 +(a−b)^2 ((1/(cos^2 ((x/2))))−1)})) =((a^2 −b^2 )/(2 cos^2 ((x/2)){(a+b)^2 cos^2 ((x/2)) +(a−b)^2 (1−cos^2 ((x/2)))(1/(cos^2 ((x/2)))))) = ((a^2 −b^2 )/(2{ ( (a+b)^2 −(a−b^2 ))cos^2 ((x/2)) +(a−b)^2 })) =((a^2 −b^2 )/(2{ 4ab cos^2 ((x/2)) +(a−b)^2 })) =((a^2 −b^2 )/(4ab(1+cosx) +2(a^2 −2ab +b^2 ))) =((a^2 −b^2 )/(4ab cosx +2a^2 +2b^2 )) = ((a^2 −b^2 )/(2(a^2 +b^2 +2ab cosx))) ⇒ y^((1)) (x) =(2/(√(a^2 −b^2 )))w^′ (x) = ((√(a^2 −b^2 ))/(a^2 +b^2 +2ab cosx)) so something went wrong in the question...$
$i f y_{1} m e a n s d e r i v a t i v e l e t w (x) = \frac{\sqrt{a^{2} - b^{2}}}{2} y (x) \Rightarrow w (x) = a r c t a n (\frac{a - b}{a + b} t a n (\frac{x}{2}))$ $\Rightarrow w^{^{'}} (x) = \frac{\frac{a - b}{2 (a + b)} {1 + {t a n}^{2} (\frac{x}{2})}}{1 + \frac{{(a - b)}^{2}}{{(a + b)}^{2}} {t a n}^{2} (\frac{x}{2})} = \frac{(a - b)}{2 (a + b) {c o s}^{2} (\frac{x}{2}) {\frac{{(a + b)}^{2} + {(a - b)}^{2} {t a n}^{2} (\frac{x}{2})}{{(a + b)}^{2}}}}$ $= \frac{a^{2} - b^{2}}{2 {c o s}^{2} (\frac{x}{2}) {{(a + b)}^{2} + {(a - b)}^{2} (\frac{1}{{c o s}^{2} (\frac{x}{2})} - 1)}}$ $= \frac{a^{2} - b^{2}}{2 {c o s}^{2} (\frac{x}{2}) {{(a + b)}^{2} {c o s}^{2} (\frac{x}{2}) + {(a - b)}^{2} (1 - {c o s}^{2} (\frac{x}{2})) \frac{1}{{c o s}^{2} (\frac{x}{2})}}$ $= \frac{a^{2} - b^{2}}{2 {({(a + b)}^{2} - (a - b^{2})) {c o s}^{2} (\frac{x}{2}) + {(a - b)}^{2}}}$ $= \frac{a^{2} - b^{2}}{2 {4 a b {c o s}^{2} (\frac{x}{2}) + {(a - b)}^{2}}} = \frac{a^{2} - b^{2}}{4 a b (1 + c o s x) + 2 (a^{2} - 2 a b + b^{2})}$ $= \frac{a^{2} - b^{2}}{4 a b c o s x + 2 a^{2} + 2 b^{2}} = \frac{a^{2} - b^{2}}{2 (a^{2} + b^{2} + 2 a b c o s x)} \Rightarrow$ $y^{(1)} (x) = \frac{2}{\sqrt{a^{2} - b^{2}}} w^{^{'}} (x) = \frac{\sqrt{a^{2} - b^{2}}}{a^{2} + b^{2} + 2 a b c o s x} s o s o m e t h i n g w e n t w r o n g i n t h e$ $q u e s t i o n . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

$i t h i n k t h e t r u e q u e s t i o n i s$ $y = \frac{2}{\sqrt{a^{2} - b^{2}}} {t a n}^{- 1} (\sqrt{\frac{a - b}{a + b}} t a n \frac{x}{2})$ $a n d a n s w e r i s o p t i o n b y_{2} = \frac{b s i n x}{{(a + b c o s x)}^{2}}$ $w h i c h i h a v e a l r e a d y p r o v e d . . .$
Commented by Tinkutara last updated on 27/Jul/18
Thanks Sir imad.
Commented by math khazana by abdo last updated on 28/Jul/18

$y o u a r e w e l c o m e s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

	Commented by Tinkutara last updated on 25/Jul/18
	Sir no option matching?
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

	$y_{1} = \frac{1}{a + b c o s x}$ $y_{2} = - \frac{1}{{(a + b c o s x)}^{2}} \times \frac{d}{d x} (a + b c o s x)$ $= \frac{- 1}{{(a + b c o s x)}^{2}} \times (- b s i n x)$ $= \frac{b s i n x}{{(a + b c o s x)}^{2}}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

	$y_{1} = \frac{\sqrt{a^{2} - b^{2}}}{a^{2} + b^{2} + 2 a b c o s x}$ $y_{2} = (- 1) \frac{\sqrt{a^{2} - b^{2}}}{{(a^{2} + b^{2} + 2 a b c o s x)}^{2}} \times \frac{d}{d x} (a^{2} + b^{2} + 2 a b c o s x)$ $= \frac{\sqrt{a^{2} - b^{2}} \times (- 2 a b s i n x)}{{(a^{2} + b^{2} + 2 a b c o s x)}^{2}} \times (- 1)$ $= \frac{(\sqrt{a^{2} - b^{2}}) (2 a b s i n x)}{{(a^{2} + b^{2} + 2 a b c o s x)}^{2}}$
	Commented by Tinkutara last updated on 27/Jul/18
	Thank you very much Sir! I got the answer. ��
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