Inverse Tan: Definition, Formulas, Graph and Properties

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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-:

Considering the trigonometric identity
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

Read More: Value of log1 to 10

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

Read More: Negative of a Vector


Graph of Inverse Tangent

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Tan-1x is the inverse of tangent function. It is denoted by the following graph-:

Graph of Inverse Tangent
Graph of Inverse Tangent

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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) =
  • 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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CBSE CLASS XII Related Questions

  • 1.

    An NGO organises a charity event in which they decide to distribute woollen caps to protect children from winter. The caps to be distributed are in three separate boxes, Box I has 30 red caps, Box II has 20 red and 10 green caps, and Box III has 30 green caps. The probability that a Box i is selected and a cap picked out is i/6, where i = 1, 2, 3.  
    Based on the above information, answer the following questions :


      • 2.
        If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


          • 3.
            Which of the following equations is NOT a Linear Differential Equation?

              • \((1 + x^2) \, dy + 2xy \, dx = \cot x \, dx\)
              • \(y + \frac{d}{dx}(xy) = x(\sin x + \log x)\)
              • \(x(1 + y^2) \, dx - y(1 + x^2) \, dy = 0\)
              • \(y \, dx - (x + 3y^2) \, dy = 0\)

            • 4.
              Find:

              If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

                • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
                • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
                • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
                • \(p = 0, \, q = 0\)

              • 5.
                Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


                  • 6.
                    Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

                      CBSE CLASS XII Previous Year Papers

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