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The derivative of \tan ^{-1}\left ( \frac{\sin x-\cos x}{\sin x+\cos x} \right ),

with respect to \frac{x}{2}, where \left ( c\epsilon \left ( 0,\frac{\pi }{2} \right ) \right ) is :

 

  • Option 1)

    1

     

     

  • Option 2)

    \frac{2}{3}

  • Option 3)

    \frac{1}{2}

  • Option 4)

    2

 

Answers (1)

Derivation of \tan ^{-1}\left ( \frac{\sin x-\cos x}{\sin x+\cos x} \right )

y= \tan ^{-1}\left ( \frac{\sin x-\cos x}{\sin x+\cos x} \right )

y= \tan ^{-1}\left ( \frac{\tan x-1}{\tan x+1}\right )

\Rightarrow y= \tan ^{-1}\left ( \tan \left ( x-\frac{\pi }{4} \right ) \right )        \left \{ Given x\epsilon \left ( 0,\frac{\pi }{2} \right ) \right \}

\Rightarrow y=x-\frac{\pi }{4}

Now, \frac{dy}{d(\frac{x}{2})}=2\frac{dy}{dx}=2


Option 1)

1

 

 

Option 2)

\frac{2}{3}

Option 3)

\frac{1}{2}

Option 4)

2

Posted by

Vakul

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