Question

If
$\tan(y) = \frac{ \sqrt{1 + \sin(x) } + \sqrt{1 - \sin(x) } }{ \sqrt{1 + \sin(x) - \sqrt{1 - \sin(x) } } }$
find dy/dx.​

Answer 1

Answer:

$\small{ \rm{tan \: y = \frac{ \sqrt{1 + sinx} + \sqrt{1 - sinx} }{ \sqrt{1 + sinx} - \sqrt{1 - sinx} } }}$

Taking LHS

To find

dy/dx

Using

sinθ = 2sinθ/2cosθ/2

$\small{ \rm{ tan \: y = \frac{ \sqrt{1 + 2sin \frac{x}{2}cos \frac{x}{2} } + \sqrt{1 + 2sin \frac{x}{2}cos \frac{x}{2} } }{\sqrt{1 + 2sin \frac{x}{2}cos \frac{x}{2} } - \sqrt{1 + 2sin \frac{x}{2}cos \frac{x}{2} }} }}$

We know that

$\small{ \rm{1 + 2sin \frac{x}{2} cos \frac{x}{2} }} \\ = \small{ \rm{ {sin}^{2} \frac{x}{2} + {cos}^{2} \frac{x}{2} + 2sin \frac{x}{2} cos \frac{x}{2}}} \\ { \rm{ = (sin \frac{x}{2} + \cos \frac{x}{2} ) ^{2} }} \\ \small { \rm{similarly \: }} \\ \small{ \rm{1 - 2sin \frac{x}{2}cos \frac{x}{2} = ( {sin} \frac{x}{2} - {cos} \frac{x}{2} ) ^{2} }}$

Substituting we get

${ \rm{ = \frac{ \sqrt{ {(sin \frac{x}{2} + cos \frac{x}{2} )}^{2} } + \sqrt{ {(sin \frac{x}{2} - cos \frac{x}{2} ) }^{2} } }{\sqrt{ {(sin \frac{x}{2} + cos \frac{x}{2} )}^{2} } - \sqrt{ {(sin \frac{x}{2} - cos \frac{x}{2} ) }^{2} } } }}$

$\small{ \rm{ = \frac{sin \frac{x}{2} + \cancel{ cos \frac{x}{2}} + sin \frac{x}{2} \cancel{- cos \frac{x}{2} } }{ \cancel{sin \frac{x}{2}} +cos \frac{x}{2} \cancel{- sin \frac{x}{2}} + cos \frac{x}{2} } }}$

$\small{ \rm{ \frac{ \cancel2sin \frac{x}{2} }{ \cancel{2}cos \frac{x}{2} } = \frac{sin \frac{x}{2} }{cos \frac{x}{2} } }}$

$\small{ \rm{tan \: y = tan \frac{x}{2} }}$

${ \rm{\frac{dtan \: y}{dx} = \frac{dtan \frac{x}{2} }{dx} }}$

${ \rm{ {sec}^{2} y \frac{dy}{dx} = {sec}^{2} \frac{x}{2} \times \frac{1}{2} }}$

${ \rm{\frac{dy}{dx} = \frac{ {sec}^{2} \frac{x}{2} }{2 {sec}^{2}y } = \frac{ 2{cos}^{2} y}{ {cos}^{2} \frac{x}{2} } }}$