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

• General and Principal Values of sin$^{-1}$ x
• General and Principal Values of cos$^{-1}$ x
• General and Principal Values of tan$^{-1}$ x
• General and Principal Values of csc$^{-1}$ x
• General and Principal Values of sec$^{-1}$ x
• General and Principal Values of cot$^{-1}$ x
• Principal Values of Inverse Trigonometric Functions
• General Values of Inverse Trigonometric Functions
  
• arcsin(x) + arccos(x) = $\frac{π}{2}$
• arctan(x) + arccot(x) = $\frac{π}{2}$
• arctan(x) + arctan(y) = arctan($\frac{x + y}{1 - xy}$)
• arctan(x) - arctan(y) = arctan($\frac{x - y}{1 + xy}$)
• arctan(x) + arctan(y) + arctan(z)= arctan$\frac{x + y + z – xyz}{1 – xy – yz – zx}$
• arccot(x) + arccot(y) = arccot($\frac{xy - 1}{y + x}$)
• arccot(x) - arccot(y) = arccot($\frac{xy + 1}{y - x}$)
• arcsin(x) + arcsin(y) = arcsin(x $\sqrt{1 - y^{2}}$ + y$\sqrt{1 - x^{2}}$)
• arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y$\sqrt{1 - x^{2}}$)
• arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• arccos(x) - arccos(y) = arccos(xy + $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• 2 arcsin(x) = arcsin(2x$\sqrt{1 - x^{2}}$) 
• 2 arccos(x) = arccos(2x$^{2}$ - 1)
• 2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$)
• 3 arcsin(x) = arcsin(3x - 4x$^{3}$)
• 3 arccos(x) = arccos(4x$^{3}$ - 3x)
• 3 arctan(x) = arctan($\frac{3x - x^{3}}{1 - 3 x^{2}}$)
• Inverse Trigonometric Function Formula
• Principal Values of Inverse Trigonometric Functions
• Problems on Inverse Trigonometric Function

11 and 12 Grade Math

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