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.






11 and 12 Grade Math

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