Question

$y = { \sin }^{ - 1} ( \frac{a + b \cos(x) }{b + a \cos(x) } ) \\ then \: \frac{dy}{dx} =$
(a)
$\frac{ - \sqrt{ {b}^{2} - {a}^{2} } }{b + a \cos(x) }$
(b)
$\frac{ - \sqrt{ {b}^{2} - {a}^{2} } }{a + b \cos(x) }$
(c)
$\frac{ - \sqrt{ {b}^{2} - {a}^{2} } \sin(x) }{ {(a + b \cos(x)) }^{2} }$
(d)
$none \: of \: these$
bahut mehnat ki hai type karne mein plz give only correct answer​

Answer 1

$y = {sin}^{ - 1} ( \frac{a + bcosx}{b + acosx} ) \\ \frac{dy}{dx} = \frac{1}{ \sqrt{1 - ( { \frac{a + bcosx}{b + acosx} })^{2} } } \frac{(b + acosx)( - bsinx) - (a + bcosx)( - asinx)}{( {b + acosx})^{2} } \\ = \frac{b + acosx}{ \sqrt{ {(b + acosx})^{2} - ( {a + bcosx})^{2} } } \frac{ - {b}{^2}sinx - absinxcosx + {a}^{2}sinx + absinxcosx }{( {b + acosx})^{2} } \\ = \frac{1}{ \sqrt{ {b}^{2} - {a}^{2} + {a}^{2} cos ^{2} x - {b}^{2}cos ^{2} x} } \frac{sinx( {a}^{2} - {b}^{2} )}{b + acosx} \\ = \frac{1}{ \sqrt{ {b}^{2} - {a}^{2} + ( {a}^{2} - {b}^{2} )(1 - {sin}^{2} x)} } \frac{ - sinx( {b}^{2} - {a}^{2}) }{b + acosx} \\ = \frac{1}{ \sqrt{ {b}^{2} - {a}^{2} + {a}^{2} - {a}^{2} {sin}^{2} x - {b}^{2} + {b}^{2} {sin}^{2} x} } \frac{ - sinx( {b}^{2} - {a}^{2}) }{b + acosx} \\ = \frac{1}{sinx \sqrt{ {b}^{2} - {a}^{2} } } \frac{ - sinx( {b}^{2} - {a}^{2} )}{b + acosx} \\ = \frac{ - ( {b}^{2} - {a}^{2} )}{ \sqrt{ {b}^{2} - {a}^{2} } } \frac{1}{b + acosx} \\ ➡ \frac{ - \sqrt{ {b}^{2} - {a}^{2} } }{b + acosx}$

Answer 2

Answer:

hi Harsh!!

am Rachana..

I am fine ..wau?