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.







11 and 12 Grade Math

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