Math, asked by Ezee, 11 hours ago

Find the derivative of tan^-1{x/1+√(1-x²)} w.r.t. sin[2 cot^-1√(1+x)/√(1-x)]

Answers

Answered by mathdude500
2

\large\underline{\sf{Solution-}}

Let we assume that

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{x}{1 +  \sqrt{1 -  {x}^{2} } } \bigg)

and

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + x} }{ \sqrt{1 - x} } \bigg)  \bigg\}

Consider,

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{x}{1 +  \sqrt{1 -  {x}^{2} } } \bigg)

Put x = siny, we get

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{siny}{1 +  \sqrt{1 -  {sin}^{2}y } } \bigg)

We know,

 \red{ \boxed{ \rm{ \:  {sin}^{2}x +  {cos}^{2}x = 1}}}

So, using this

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{siny}{1 +  \sqrt{{cos}^{2}y } } \bigg)

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{siny}{1 + {{cos}y } } \bigg)

We know,

\red{ \boxed{ \rm{ \: 1 + cos2x =  {2cos}^{2}x}}}

and

\red{ \boxed{ \rm{ \: 1  -  cos2x =  {2sin}^{2}x}}}

So, using this identity, we get

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{2sin\dfrac{y}{2} cos\dfrac{y}{2} }{2 {cos}^{2}\dfrac{y}{2}  } \bigg)

\rm :\longmapsto\:u =  {tan}^{ - 1}\bigg(\dfrac{sin\dfrac{y}{2}  }{ {cos}\dfrac{y}{2}  } \bigg)

\rm :\longmapsto\:u =  {tan}^{ - 1}tan\dfrac{y}{2}

\rm :\longmapsto\:u = \dfrac{y}{2}

\rm :\longmapsto\:u = \dfrac{1}{2}  {sin}^{ - 1}x

On differentiating both sides, w. r. t. x, we get

\rm :\longmapsto\:\dfrac{d}{dx}u =\dfrac{d}{dx} \dfrac{1}{2}  {sin}^{ - 1}x

\rm :\longmapsto\:\dfrac{du}{dx} = \dfrac{1}{2}  \dfrac{d}{dx}{sin}^{ - 1}x

We know,

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {sin}^{ - 1}x =  \frac{1}{ \sqrt{1 -  {x}^{2} } }}}}

So, using this, we get

\bf :\longmapsto\:\dfrac{du}{dx} = \dfrac{1}{2}  \dfrac{1}{ \sqrt{1 -  {x}^{2} } } -  -  - (1)

Consider,

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + x} }{ \sqrt{1 - x} } \bigg)  \bigg\}

Put x = cosy, we get

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{1 + cosy} }{ \sqrt{1 - cosy} } \bigg)  \bigg\}

We know,

\red{ \boxed{ \rm{ \: 1 + cos2x =  {2cos}^{2}x}}}

and

\red{ \boxed{ \rm{ \: 1  -  cos2x =  {2sin}^{2}x}}}

So, on using these Identities, we get

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ \sqrt{2cos ^{2}  \dfrac{y}{2} } }{ \sqrt{2sin ^{2}  \dfrac{y}{2} } } \bigg)  \bigg\}

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(\dfrac{ cos \dfrac{y}{2} }{ sin\dfrac{y}{2} } \bigg)  \bigg\}

\rm :\longmapsto\:v =  sin \:  \bigg \{2{cot}^{ - 1}\bigg(cot \dfrac{y}{2} \bigg)  \bigg\}

\rm :\longmapsto\:v =  sin \:  \bigg \{2\bigg(\dfrac{y}{2} \bigg)  \bigg\}

\rm :\longmapsto\:v = siny

We know,

 \red{ \boxed{ \rm{ \:  {sin}^{2}x +  {cos}^{2}x = 1}}}

So, using this

\rm :\longmapsto\:v =  \sqrt{1 -  {cos}^{2} y}

\rm :\longmapsto\:v =  \sqrt{1 -  {x}^{2}}

On differentiating both sides w. r. t. x, we get

\rm :\longmapsto\:\dfrac{d}{dx}v = \dfrac{d}{dx} \sqrt{1 -  {x}^{2}}

We know,

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} \sqrt{x} =  \frac{1}{2 \sqrt{x} }}}}

So, using this

\rm :\longmapsto\:\dfrac{dv}{dx} = \dfrac{1}{2 \sqrt{1 -  {x}^{2} } } \dfrac{d}{dx}(1 -  {x}^{2})

\rm :\longmapsto\:\dfrac{dv}{dx} = \dfrac{1}{2 \sqrt{1 -  {x}^{2} } }( - 2x)

\rm :\longmapsto\:\dfrac{dv}{dx} = -  \:  \dfrac{x}{\sqrt{1 -  {x}^{2} } } -  -  - (2)

Now, Consider

\rm :\longmapsto\:\dfrac{du}{dv}

\rm \:  =  \:  \: \dfrac{du}{dx} \:  \div  \: \dfrac{dv}{dx}

\rm \:  =  \:  \: \dfrac{1}{2 \sqrt{1 -  {x}^{2} } } \:  \div  \: \dfrac{ - x}{ \sqrt{1 -  {x}^{2} } }

\rm \:  =  \:  \:  -  \: \dfrac{1}{2x}

Additional Information :-

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {cos}^{ - 1}x =  \frac{ - 1}{ \sqrt{1 -  {x}^{2} } }}}}

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {tan}^{ - 1}x =  \frac{1}{1 +  {x}^{2} }}}}

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {cot}^{ - 1}x =  \frac{ -  \: 1}{1 +  {x}^{2} }}}}

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {sec}^{ - 1}x =  \frac{1}{x \sqrt{ {x}^{2}  - 1} } }}}

\red{ \boxed{ \rm{ \: \dfrac{d}{dx} {cosec}^{ - 1}x =  \frac{ -  \: 1}{x \sqrt{ {x}^{2}  - 1} } }}}

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