# 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.

Inverse Trigonometric Functions

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