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3 arctan(x) = arctan(3x−x31−3x2)
We will learn how to prove the property of the inverse trigonometric function 3 arctan(x) = arctan(3x−x31−3x2) or, 3 tan−1 x = tan−1 (3x−x31−3x2).
Proof:
Let, tan−1 x = θ
Therefore, tan θ = x
Now we know that, tan 3θ = 3tanθ−tan3θ1−3tan2θ
⇒ tan 3θ = 3x−x31−3x2
Therefore, 3θ = tan−1 (3x−x31−3x2)
⇒ 3 tan−1 x = tan−1 (3x−x31−3x2)
or, 3 arctan(x) = arctan(3x−x31−3x2). Proved.
● 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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