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\(^{-1}\) \(\frac{56}{33}\).

Solution:  

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\(^{-1}\) \(\frac{4}{3}\)

and

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.







11 and 12 Grade Math

From arctan x + arccot x = π/2 to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.