Math, asked by someyuck, 9 months ago

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

Answers

Answered by ITzBrainlyGuy
15

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} }  }}

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