# 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

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