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Inverse Tangent is the inverse function of the trigonometric function Tangent. Inverse tangent is generally denoted by tan-1 (x). Tangent is the ratio of perpendicular and base of a triangle. All the inverse trigonometric functions are also known as arcus functions, so tan-1 (x) can also be written as arctan. Inverse Trigonometric Functions are an integral part of Calculus. The formulas are used in problems based on differentiation and integration. These functions are useful in construction, architecture, engineering, cartography etc.
Read More: Applications of Trigonometry
Key Terms: Inverse Tangen, Functions, Perpendicular, Base of a triangle, Tangent, Differentiation, Integration, Triangle, Angle
Read More: Addition And Subtraction Of Integers
Definition of Inverse Tan
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Tangent function has an inverse Tan-1 (x)that works in reverse. It is used to determine the angle when perpendicular and base of a triangle is given. The Derivative of tan-1x is 1/(1 + x2) is used in solving numerous questions. The function can be integrated using integration by parts. With this formula we can find the value of an angle when the value of tangent is given.
Read More: Ratios and Identities in Trigonometry
Formulas of Inverse Tan
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i) Considering the trigonometric identity-:
When we put a = tan-1x ,b = tan-1y, we get
- tan-1x + tan-1y = tan-1(x + y)/(1 - xy)
- tan-1x - tan-1y = tan-1(x - y)/(1 + xy)
Inverse Tan formulas for Integration
- ∫tan-1x dx = {x tan-1x (x)} – {Iog (x2+1)/2} + C
- ∫tan-1 (ax) dx = {x tan-1(ax)}- {Iog(a2x2+1)/2a} + C
- ∫x.tan-1(ax)dx = {x2 tan-1(ax)/2} – {tan-1(ax)/2a2 } – {x / 2a} + C
- ∫ x2 tan-1(ax)dx = {x3 tan-1(ax)/3} – {Iog (a2x2 + 1)/6a3 } – {x2/6a} + C
Read More: Definite Integral Formula
Formulas of Inverse Tangent Function in Calculus
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Inverse Tangent Function is an important part of calculus. It is useful in solving questions based on derivation and integration. The derivative and integral of tan-1x appear throughout the syllabus of mathematics. It is used in calculus in the following ways-:
i) Inverse Tangent Function in Derivation
Inverse tangent is used to solve the problems based on derivation. The derivative of tan-1x is denoted by d.tan-1x /dx. The derivative of tan-1x comes out to be 1/1+x2 . The derivative of tan-1x is useful for defining the graph plotted between π/2 and –π/2.
ii) Inverse Tangent Function in Integration
Inverse Tangent Function is used in solving problems based on integration.There is no direct way to integrate tan-1x. However, tan-1x can be integrated using integration by parts. The integration of tan-1x can be calculated as follows-:
So, now we need to integrate tan-1x.
We know that d(tan-1x)/dx = 1/(1 + x2)
We will use the formula ∫uv dx = u ∫vdx - ∫[du/dx . ∫vdx] dx
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Here u = tan-1x, v = 1
∫ tan-1x dx = ∫ (tan-1x × 1) dx
= x tan-1x – ∫x/(1+x2) dx
= x tan-1x – ½ ∫2x/(1+x2) dx ----- (1)
Now, assume 1 + x2 = t ⇒ 2x dx = dt
∫2x/(1+x2) dx = ∫1/t dt = log |t| + C = log|1 + x2| + C
Substitute ∫2x/(1+x2) dx = log|1 + x2| + C in (1)
∫ tan-1x dx = x tan-1x - ½ log|1+x2| + C
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Graph of Inverse Tangent
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Tan-1x is the inverse of tangent function. It is denoted by the following graph-:
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Relationship between Inverse Tan and other Trigonometric Functions
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The relationship between inverse tan and different trigonometric identities. Trigonometric identities are generally applied in a right angled triangle. To define the relationship, we take perpendicular as x, base as 1 and hypotenuse as√(1 - x2)
1+x2
The relationship between Inverse Tan and Trigonometric Identities can be defined as -:
- Sin (tan-1 x) = x/ (√1+ x2)
- Cos (tan-1 x) = 1 / (√1+ x2)
- Tan (tan-1 x) = x
Also Read: Decimal Expansion of Rational Numbers with Solved Examples
Properties of Inverse Tan
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Some of the important properties of Inverse Tan are-:
- Annotation: tan-1x
- Definition: x = Tan(x)
- Domain: All Real Numbers
- Range: -π/2 < y < π/2
Read More: Inverse Trigonometric Formula
mportant Values of Inverse Tan
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The value of Inverse tan can vary depending upon the angle. Some of the important values of Inverse Trigonometry are as follows-:
- Tan-1 (0) = 0°
- Tan-1 (1/√3) = 30°
- Tan-1 (1) = 45°
- Tan-1 (√3) = 60°
- Tan-1 (∞) = 90°
Read More: Introduction to Trigonometry: Ratios, Identities & Complementary Angle
Things to Remember
- Inverse Tangent is the inverse of Tangent function. It is denoted by tan-1x. It is also known as arctan as it is an arcus function.
- Inverse Tangent is used in engineering, architecture, cartography, marine biology etc.
- Some of important formulas of inverse tangent are-:
- tan-1x + tan-1y = tan-1(x + y)/(1 - xy)
- tan-1x - tan-1y = tan-1(x - y)/(1 + xy)
- Inverse Tangent is very important in calculus. It is used to solve problems based on integration and differentiation.
- The derivative of tan-1x is 1/(1 + x2).
- The integral of tan-1x is {x tan-1x (x)} – {Iog (x2+1)/2} + C.
- The domain of tan-1x is All Real Numbers.
- The range of tan-1x -π/2 < y < π/2.
Also Read:
Sample Questions
Ques. Prove that: Tan- 1 63/16 = Sin- 1 5/13 + Cos- 1 3/5. (5 marks)
Ans. Let sin- 1 5/13 = x
⇒ sin x = 5/13
Then, cosx = √1 - (5/13)2 = 12/13
Therefore,
tan x = 5/12
x = tan- 1 5/12
sin-1 5/13 = tan- 1 5/12 ....(1)
Now, let cos- 1 3/5 = y
⇒ cos y = 3/5
Then, sin y = √1 - (3/5)2
= √16/25
= 4/5
Therefore, tan y = 4/3
y = tan- 1 4/3
cos-1 3/5 = tan- 1 4/3 ....(2)
Thus, by using (1) and (2)
LHS = sin- 1 5/12 + cos- 1 3/5
= tan- 1 5/12 + tan- 1 4/3
= tan- 1 [(5/12 + 4/3)/(1 - 5/12.4/3)]
= tan- 1 [((32 + 45)/60/(60 - 24)/60]
= tan- 1 63/16
= LHS
Ques. Find the Value of: Tan-1 x/√(a2 – x2). (2 marks)
Ans. Let y = tan-1 x/√(a2 – x2)
Put x = a sin θ
θ = sin-1(x/a)
So y = tan-1 (a sin θ/√(a2 – a2 sin2 θ))
= tan-1 (a sin θ/a√(1 – sin2 θ))
= tan-1 (a sin θ/a√cos 2 θ)
= tan-1 (a sin θ/a cos θ)
= tan-1 tan θ
= θ
= sin-1(x/a)
Ques. Find the value: cot(tan-1x +cot-1x). (2 marks)
Ans. Given function is cot( tan-1x+ cot-1x)
We know that, tan-1x + cot-1x = π/2
So, we get cot(π/2)
= 0
Ques. Find the value of tan-1(tan3π/4). (2 marks)
Ans. The principle of value of tan-1x is (-π/2,π/2)
So, tan-1(tan3π/4) = tan-1(tan {π-π/4})
= tan-1(tan{-π/4})
= -π/4
Ques. Find the value of : tan-1(√3) - cot-1(-√3). (2 marks)
Ans. tan-1(√3) - cot-1(-√3)
= tan-1(√3) - (π - cot-1(√3))
= tan-1(√3) - π + cot-1(√3)
= π/3 - π + π/6
= π/2 - π
= -π/2
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