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.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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