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
11 and 12 Grade Math
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