arctan(x) + arctan(y) = arctan(\(\frac{x + y}{1 - xy}\))

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arctan(y) = arctan(\(\frac{x + y}{1 - xy}\)), (i.e., tan\(^{-1}\) x + tan\(^{-1}\) y = tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\)) if x > 0, y > 0 and xy < 1.

1. Prove that arctan(x) + arctan(y) = arctan(\(\frac{x + y}{1 - xy}\)), if x > 0, y > 0 and xy < 1.

Proof:

Let, tan\(^{-1}\) x = α and tan\(^{-1}\) y = β

From tan\(^{-1}\) x = α we get,

x = tan α

and from tan\(^{-1}\) y = β we get,

y = tan β

Now, tan (α + β) = (\(\frac{tan α + tan β}{1 - tan α tan β}\))

tan (α + β) = \(\frac{x + y}{1 - xy}\)

⇒ α + β = tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\))

⇒ tan\(^{-1}\) x + tan\(^{-1}\) y = tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\))

Therefore, tan\(^{-1}\) x + tan\(^{-1}\) y = tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\)), if x > 0, y > 0 and xy < 1.


2. Prove that arctan(x) + arctan(y) = π + arctan(\(\frac{x + y}{1 - xy}\)), if x > 0, y > 0 and xy > 1. And

arctan(x) + arctan(y) =  arctan(\(\frac{x + y}{1 - xy}\)) - π, if x < 0, y < 0 and xy > 1.

Proof: If x > 0, y > 0 such that xy > 1, then \(\frac{x + y}{1 - xy}\) is positive and therefore, \(\frac{x + y}{1 - xy}\) is positive angle between 0° and 90°.

Similarly, if x < 0, y < 0 such that xy > 1, then \(\frac{x + y}{1 - xy}\) is positive and therefore, tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\)) is a negative angle while tan\(^{-1}\) x + tan\(^{-1}\) y is a positive angle while tan\(^{-1}\) x + tan\(^{-1}\) y is a non-negative angle. Therefore, tan\(^{-1}\)  x + tan\(^{-1}\) y = π + tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\)), if x > 0, y > 0 and xy > 1 and

arctan(x) + arctan(y) =  arctan(\(\frac{x + y}{1 - xy}\)) - π, if x < 0, y < 0 and xy > 1.


Solved examples on property of inverse circular function tan\(^{-1}\) x + tan\(^{-1}\) y = tan\(^{-1}\) (\(\frac{x + y}{1 - xy}\))

1. Prove that 4 (2 tan\(^{-1}\) \(\frac{1}{3}\) + tan\(^{-1}\) \(\frac{1}{7}\)) = π

Solution:  

2 tan\(^{-1}\) \(\frac{1}{3}\)

= tan\(^{-1}\) \(\frac{1}{3}\) + tan\(^{-1}\) \(\frac{1}{3}\)

= tan\(^{-1}\) (\(\frac{\frac{1}{3} + \frac{1}{3}}{1 - \frac{1}{3} • \frac{1}{3}}\))

= tan\(^{-1}\) \(\frac{3}{4}\)

Now L. H. S. = 4 (2 tan\(^{-1}\) \(\frac{1}{3}\) + tan\(^{-1}\) \(\frac{1}{7}\))

= 4 (tan\(^{-1}\) \(\frac{3}{4}\) + tan\(^{-1}\) \(\frac{1}{7}\))

= 4 tan\(^{-1}\) (\(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} • \frac{1}{7}}\))

= 4 tan\(^{-1}\) (\(\frac{25}{28}\) x \(\frac{28}{25}\))

= 4 tan\(^{-1}\) 1

= 4 · \(\frac{π}{4}\)

= π = R.H.S.                        Proved.


2. Prove that, tan\(^{-1}\) \(\frac{1}{4}\) + tan\(^{-1}\) \(\frac{2}{9}\) + tan\(^{-1}\) \(\frac{1}{5}\) + tan\(^{-1}\) \(\frac{1}{8}\) = π/4.

Solution:

L. H. S. = tan\(^{-1}\) \(\frac{1}{4}\) + tan\(^{-1}\) \(\frac{2}{9}\) + tan\(^{-1}\) \(\frac{1}{5}\) + tan\(^{-1}\) \(\frac{1}{8}\)

= tan\(^{-1}\) \(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} • \frac{2}{9}}\) + tan\(^{-1}\) \(\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} • \frac{1}{8}}\)

= tan\(^{-1}\) (\(\frac{17}{36}\) x \(\frac{36}{34}\)) + tan\(^{-1}\) (\(\frac{13}{40}\) x \(\frac{40}{39}\))

= tan\(^{-1}\) \(\frac{1}{2}\) + tan\(^{-1}\) \(\frac{1}{3}\)

= tan\(^{-1}\) \(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} • \frac{1}{3}}\)

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

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

 Inverse Trigonometric Functions


11 and 12 Grade Math

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