Question

what is the formula os 2 tan inverse x

Answer 1

formula of 2 tan inverse X equal to tan inverse 2 X by 1 - x square

Answer 2

Answer:

There are 3 formulas of $2tan^{-1}x$ ,

$2tan^{-1}x$ =  $sin^{-1} \frac{2x}{(1+x^{2} )}$, |x| ≤ 1

$2tan^{-1}x$ = $cos^{-1} \frac{1-x^{2} }{(1+x^{2} )}$, x ≥ 0

$2tan^{-1}x$ = $tan^{-1} \frac{2x}{(1-x^{2} )}$, -1<x<1

Step-by-step explanation:

Step 1:- Let, $tan^{-1}$x = θ ------ ( 1 )

            and tan θ = a,   ------ ( 2 ) [ I took "a" for a different alphabet you can take any]

Step 2:- As we know,

tan 2θ = $[\frac{2tan\theta}{1-tan^{2}\theta } ]$

tan 2θ = $[\frac{2a}{1-a^{2} } ]$        [ replacing tan θ with a, from ( 1 ) ]

2θ = $tan^{-1}$$[\frac{2a}{1-a^{2} } ]$      [ we took "tan" on the other side of the equals resulting in "$tan^{-1}$" ]

2  $tan^{-1}$x = $tan^{-1}$$[\frac{2a}{1-a^{2} } ]$ ------------ ( i )

Step 3:- Again,

sin 2θ = $[\frac{2tan\theta}{1+tan^{2}\theta } ]$

sin 2θ = $[\frac{2a}{1+a^{2} } ]$          [ replacing tan θ with a, from ( 1 ) ]

2θ = $sin^{-1}$$[\frac{2a}{1+a^{2} } ]$       [ we took "sin" on the other side of the equals resulting in "$sin^{-1}$" ]

2  $tan^{-1}$x = $sin^{-1}$$[\frac{2a}{1+a^{2} } ]$ ------------ ( ii )

Step 4:- Now,

cos 2θ = $[\frac{1-tan^{2}\theta}{1+tan^{2}\theta } ]$

cos 2θ = $[\frac{1-a^{2}}{1+a^{2} } ]$     [ replacing tan θ with a, from ( 1 ) ]

2θ = $cos^{-1}$$[\frac{1-a^{2}}{1+a^{2} } ]$    [ we took "cos" on the other side of the equals resulting in " $cos^{-1}$" ]

2  $tan^{-1}$x = $cos^{-1}$$[\frac{1-a^{2}}{1+a^{2} } ]$ ---------- ( iii )

Hence proved all three formulas.

To know more about such questions,

Click here:- https://brainly.in/question/5776077

Click here:- https://brainly.in/question/6742614