# arccot(x) + arccot(y) = arccot($$\frac{xy - 1}{y + x}$$)

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

Proof:

Let, cot$$^{-1}$$ x = α and cot$$^{-1}$$ y = β

From cot$$^{-1}$$ x = α we get,

x = cot α

and from cot$$^{-1}$$ y = β we get,

y = cot β

Now, cot (α + β) = ($$\frac{cot α cot β - 1}{cot β + tan α}$$)

cot (α + β) = $$\frac{xy - 1}{y + x}$$

⇒ α + β = cot$$^{-1}$$ $$\frac{xy - 1}{y + x}$$

⇒ cot$$^{-1}$$ x + cot$$^{-1}$$ y = cot$$^{-1}$$ $$\frac{xy - 1}{y + x}$$

Therefore, cot$$^{-1}$$ x + cot$$^{-1}$$ y = cot$$^{-1}$$ $$\frac{xy - 1}{y + x}$$

Inverse Trigonometric Functions

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