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.

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Inverse Trigonometric Functions