arctan(x) + arccot(x) = \(\frac{π}{2}\)

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arccot(x) = \(\frac{π}{2}\) (i.e., tan\(^{-1}\) x + cot\(^{-1}\) x = \(\frac{π}{2}\)).

Proof: Let, tan\(^{-1}\) x = θ              

Therefore, x = tan θ

x = cot (\(\frac{π}{2}\) - θ), [Since, cot (\(\frac{π}{2}\) - θ) = tan θ]

⇒ cot\(^{-1}\) x = \(\frac{π}{2}\) - θ

⇒ cot\(^{-1}\) x= \(\frac{π}{2}\) - tan\(^{-1}\) x, [Since, θ = tan\(^{-1}\) x]

⇒ cot\(^{-1}\) x + tan\(^{-1}\) x = \(\frac{π}{2}\)

⇒ tan\(^{-1}\) x + cot\(^{-1}\) x = \(\frac{π}{2}\)

Therefore, tan\(^{-1}\) x + cot\(^{-1}\) x = \(\frac{π}{2}\).             Proved.

Solved examples on property of inverse circular function tan\(^{-1}\) x + cot\(^{-1}\) x = \(\frac{π}{2}\)

Prove that, tan\(^{-1}\) 4/3 + tan\(^{-1}\) 12/5 = π - tan\(^{-1}\) \(\frac{56}{33}\).


We know that tan\(^{-1}\) x + cot\(^{-1}\) x = \(\frac{π}{2}\)

⇒ tan\(^{-1}\) x = \(\frac{π}{2}\) - cot\(^{-1}\) x

⇒ tan\(^{-1}\) \(\frac{4}{3}\)  = \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{4}{3}\)


tan\(^{-1}\) \(\frac{12}{5}\) = \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{12}{5}\)

Now, L. H. S. = tan\(^{-1}\) \(\frac{4}{3}\) + tan\(^{-1}\) \(\frac{12}{5}\)

= \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{4}{3}\) + \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{12}{5}\), [Since, tan\(^{-1}\) \(\frac{4}{3}\) = \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{4}{3}\) and tan\(^{-1}\) \(\frac{12}{5}\) = \(\frac{π}{2}\) - cot\(^{-1}\) \(\frac{12}{5}\)]

= π - (cot\(^{-1}\) \(\frac{4}{3}\) + cot\(^{-1}\) \(\frac{12}{5}\))

= π - (tan\(^{-1}\) \(\frac{3}{4}\) + tan\(^{-1}\) \(\frac{5}{12}\))

= π – tan\(^{-1}\) \(\frac{\frac{3}{4} + \frac{5}{12}}{1 – \frac{3}{4} · \frac{5}{12}}\)

= π – tan\(^{-1}\) (\(\frac{14}{12}\) x \(\frac{48}{33}\))

= π – tan\(^{-1}\) \(\frac{56}{33}\) = R. H. S.       Proved.

 Inverse Trigonometric Functions

11 and 12 Grade Math

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