2 arctan(x) = arctan(\(\frac{2x}{1 - x^{2}}\)) = arcsin(\(\frac{2x}{1 + x^{2}}\)) = arccos(\(\frac{1 - x^{2}}{1 + x^{2}}\))

We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan(\(\frac{2x}{1 - x^{2}}\)) = arcsin(\(\frac{2x}{1 + x^{2}}\)) = arccos(\(\frac{1 - x^{2}}{1 + x^{2}}\))

or, 2 tan\(^{-1}\) x = tan\(^{-1}\) (\(\frac{2x}{1 - x^{2}}\)) = sin\(^{-1}\) (\(\frac{2x}{1 + x^{2}}\)) = cos\(^{-1}\) (\(\frac{1 - x^{2}}{1 + x^{2}}\))

Proof: 

Let, tan\(^{-1}\) x = θ         

Therefore, tan θ = x

We know that,

tan 2θ = \(\frac{2 tan θ}{1 - tan^{2}θ}\)

tan 2θ = \(\frac{2x}{1 - x^{2}}\)

2θ = tan\(^{-1}\)(\(\frac{2x}{1 - x^{2}}\))

2 tan\(^{-1}\) x = tan\(^{-1}\)(\(\frac{2x}{1 - x^{2}}\)) …………………….. (i)

Again, sin 2θ = \(\frac{2 tan θ}{1 + tan^{2}θ}\)

sin 2θ = \(\frac{2x}{1 + x^{2}}\)

2θ = sin\(^{-1}\)(\(\frac{2x}{1 + x^{2}}\) )

2 tan\(^{-1}\) x = sin\(^{-1}\)(\(\frac{2x}{1 + x^{2}}\)) …………………….. (ii)

Now,  cos 2θ = \(\frac{1 - tan^{2}θ}{1 + tan^{2}θ}\)

 cos 2θ =  \(\frac{1 - x^{2} }{1 + x^{2} }\)

2θ = cos\(^{-1}\) (\(\frac{1 - x^{2} }{1 + x^{2} }\))

2 tan\(^{-1}\) x = cos (\(\frac{1 - x^{2} }{1 + x^{2} }\)) …………………….. (iii)

Therefore, from (i), (ii) and (iii) we get, 2 tan\(^{-1}\) x = tan\(^{-1}\) \(\frac{2x}{1 - x^{2}}\) = sin\(^{-1}\) \(\frac{2x}{1 + x^{2}}\) = cos\(^{-1}\) \(\frac{1 - x^{2}}{1 + x^{2}}\)                   Proved.

 

Solved examples on property of inverse circular function 2 arctan(x) = arctan(\(\frac{2x}{1 - x^{2}}\)) = arcsin(\(\frac{2x}{1 + x^{2}}\)) = arccos(\(\frac{1 - x^{2}}{1 + x^{2}}\)):

1. Find the value of the inverse function tan(2 tan\(^{-1}\) \(\frac{1}{5}\)).

Solution:

tan (2 tan\(^{-1}\) \(\frac{1}{5}\))

= tan (tan\(^{-1}\) \(\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}\)), [Since, we know that, 2 tan\(^{-1}\) x = tan\(^{-1}\)(\(\frac{2x}{1 - x^{2}}\))]

 = tan (tan\(^{-1}\) \(\frac{\frac{2}{5}}{1 - \frac{1}{25}}\))

= tan (tan\(^{-1}\) \(\frac{5}{12}\))

= \(\frac{5}{12}\)

 

2. Prove that, 4 tan\(^{-1}\) \(\frac{1}{5}\) - tan\(^{-1}\) \(\frac{1}{70}\) + tan\(^{-1}\) \(\frac{1}{99}\) = \(\frac{π}{4}\)

Solution:

L. H. S. = 4 tan\(^{-1}\) \(\frac{1}{5}\) - tan\(^{-1}\) \(\frac{1}{70}\) + tan\(^{-1}\) \(\frac{1}{99}\)

= 2(2 tan\(^{-1}\) \(\frac{1}{5}\)) - tan\(^{-1}\) \(\frac{1}{70}\) + tan\(^{-1}\) \(\frac{1}{99}\)

= 2(tan\(^{-1}\) \(\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}\)) - tan\(^{-1}\) \(\frac{1}{70}\) + tan\(^{-1}\) \(\frac{1}{99}\), [Since, 2 tan\(^{-1}\) x = tan\(^{-1}\)(\(\frac{2x}{1 - x^{2}}\))]

= 2 (tan\(^{-1}\) \(\frac{2\frac{1}{5}}{1 - (\frac{1}{25})}\))- tan\(^{-1}\) \(\frac{1}{70}\) + tan\(^{-1}\) \(\frac{1}{99}\),

= 2 tan\(^{-1}\) \(\frac{5}{12}\) - (tan\(^{-1}\) \(\frac{1}{70}\) - tan\(^{-1}\) \(\frac{1}{99}\))

= tan\(^{-1}\) (\(\frac{2 × \frac{5}{12}}{1 - (\frac{5}{12})^{2}}\)) - tan\(^{-1}\) (\(\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{77} × \frac{1}{99}}\))

= tan\(^{-1}\) \(\frac{120}{199}\) - tan\(^{-1}\) \(\frac{29}{6931}\)

= tan\(^{-1}\) \(\frac{120}{199}\) - tan\(^{-1}\) \(\frac{1}{239}\)

= tan\(^{-1}\) (\(\frac{\frac{120}{199} - \frac{1}{239}}{1 + \frac{120}{119} × \frac{1}{239}}\))

= tan\(^{-1}\) 1

= tan\(^{-1}\) (tan \(\frac{π}{4}\))

= \(\frac{π}{4}\) = R. H. S.              Proved.

 Inverse Trigonometric Functions






11 and 12 Grade Math

From 2 arctan(x) to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

    Mar 02, 24 05:31 PM

    Fractions
    The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

    Read More

  2. Subtraction of Fractions having the Same Denominator | Like Fractions

    Mar 02, 24 04:36 PM

    Subtraction of Fractions having the Same Denominator
    To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera…

    Read More

  3. Addition of Like Fractions | Examples | Worksheet | Answer | Fractions

    Mar 02, 24 03:32 PM

    Adding Like Fractions
    To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

    Read More

  4. Comparison of Unlike Fractions | Compare Unlike Fractions | Examples

    Mar 01, 24 01:42 PM

    Comparison of Unlike Fractions
    In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…

    Read More

  5. Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples

    Feb 29, 24 05:12 PM

    Equivalent Fractions
    The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re…

    Read More