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\(^{-1}\) (\(\frac{xy + y}{y - x}\))

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\(^{-1}\) \(\frac{xy + 1}{y - x}\)











11 and 12 Grade Math

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