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arccot(x) + arccot(y) = arccot(xy−1y+x)
We will learn how to prove the property of the inverse trigonometric function arccot(x) + arccot(y) = arccot(xy−1y+x) (i.e., cot−1 x - cot−1 y = cot−1 (xy−1y+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) = (cotαcotβ−1cotβ+tanα)
cot (α + β) = xy−1y+x
⇒ α + β = cot−1 xy−1y+x
⇒ cot−1 x + cot−1 y = cot−1 xy−1y+x
Therefore, cot−1 x + cot−1 y = cot−1 xy−1y+x
● Inverse Trigonometric Functions
- General and Principal Values of sin−1 x
- General and Principal Values of cos−1 x
- General and Principal Values of tan−1 x
- General and Principal Values of csc−1 x
- General and Principal Values of sec−1 x
- General and Principal Values of cot−1 x
- Principal Values of Inverse Trigonometric Functions
- General Values of Inverse Trigonometric Functions
- arcsin(x) + arccos(x) = π2
- arctan(x) + arccot(x) = π2
- arctan(x) + arctan(y) = arctan(x+y1−xy)
- arctan(x) - arctan(y) = arctan(x−y1+xy)
- arctan(x) + arctan(y) + arctan(z)= arctanx+y+z–xyz1–xy–yz–zx
- arccot(x) + arccot(y) = arccot(xy−1y+x)
- arccot(x) - arccot(y) = arccot(xy+1y−x)
- arcsin(x) + arcsin(y) = arcsin(x √1−y2 + y√1−x2)
- arcsin (x) - arcsin(y) = arcsin (x √1−y2 - y√1−x2)
- arccos (x) + arccos(y) = arccos(xy - √1−x2√1−y2)
- arccos(x) - arccos(y) = arccos(xy + √1−x2√1−y2)
- 2 arcsin(x) = arcsin(2x√1−x2)
- 2 arccos(x) = arccos(2x2 - 1)
- 2 arctan(x) = arctan(2x1−x2) = arcsin(2x1+x2) = arccos(1−x21+x2)
- 3 arcsin(x) = arcsin(3x - 4x3)
- 3 arccos(x) = arccos(4x3 - 3x)
- 3 arctan(x) = arctan(3x−x31−3x2)
- Inverse Trigonometric Function Formula
- Principal Values of Inverse Trigonometric Functions
- Problems on Inverse Trigonometric Function
11 and 12 Grade Math
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