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$^{-1}$ ($\frac{x - y}{1 + xy}$))

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$^{-1}$ ($\frac{x - y}{1 + xy}$)

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

Solve the inverse trigonometric function: 3 tan$^{-1}$  1/2 + √3 - tan$^{-1}$ 1/x = tan$^{-1}$ 1/3

Solution:

We know that, tan 15° = tan (45° - 30°)

⇒ tan 15° = $\frac{tan 45° - tan 30°}{1 + tan 45° tan 30°}$

⇒ tan 15° = $\frac{1 - \frac{1}{√3}}{1 + \frac{1}{√3}}$

⇒ tan 15° = $\frac{√3 - 1}{√3 + 1}$

⇒ tan 15° = $\frac{(√3 - 1)(√3 + 1)}{(√3 + 1)(√3 + 1)}$

⇒ tan 15° = $\frac{3 - 1}{4 + 2√3}$

⇒ tan 15° = $\frac{1}{2 + √3}$

⇒ tan$^{-1}$ ($\frac{1}{2 + √3}$) = 15°

⇒ tan$^{-1}$ ($\frac{1}{2 + √3}$) = $\frac{π}{12}$

Therefore, from the given equation we get,

3 tan$^{-1}$ $\frac{1}{2 + √3}$ - tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ $\frac{1}{3}$

⇒ 3 · $\frac{π}{12}$ - tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ $\frac{1}{3}$

⇒ - tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ $\frac{1}{3}$ - $\frac{π}{4}$

⇒  tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ 1 - tan$^{-1}$ $\frac{1}{3}$   [Since, $\frac{π}{4}$ = tan$^{-1}$ 1]            

⇒ tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ $\frac{1 - \frac{1}{3}}{1 + 1 • \frac{1}{3}}$

⇒ tan$^{-1}$ $\frac{1}{x}$ = tan$^{-1}$ ½

⇒ $\frac{1}{x}$  = ½     

⇒ x = 2

Therefore, the required solution is x = 2.

● 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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