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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