Question

if y=sin inverse ( acosx -bsinx /bsinx + acosx ) find differential cofficient w.r.t. x.​

Answer 1

Step-by-step explanation:

We have,

$y = \sin^{ - 1} ( \frac{a \cos(x) - b \sin (x )}{a \cos(x) + b \sin(x) } )$

Now,

$\frac{dy}{dx} = \frac{d}{dx} (\frac{a \cos(x) - b \sin(x)}{a \cos(x) + b \sin(x)} ) . \frac{1}{ \sqrt{1 - { (\frac{a \cos(x) - b \sin(x) }{a \cos(x) + b \sin(x) }) }^{2} } }$

On simplifying,

$\frac{dy}{dx} = - \frac{2ab}{ {(a \cos(x) + b \sin(x ) )}^{2} } . \frac{(a \cos(x) + b \sin(x)) }{ \sqrt{4ab \sin(x) \cos(x) } }$

$\frac{dy}{dx} = - \frac{ \sqrt{2ab} }{(a \cos(x) + b \sin(x) ). \sqrt{ \sin(2x) } }$