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)
V Vakul

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

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions