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















11 and 12 Grade Math

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