Tangent 3 Theta Formula: Proof & Derivation

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Tangent 3 Theta formula is, Tan 3 theta = 3 tan theta – tan theta / 1 – 3 tanTrigonometric ratios (sin, cos, tan, sec, cosec, and tan), Pythagorean identities, and trigonometric identities are among the most commonly used ways to solve the Tan 3 theta. 

  • The period of Tan 3 Theta is Period .
  • The tangent 3 theta formula can be used at times when we simplify a triple angle tan function or expand them as per requirement. 
  • Herein, Tan is the trigonometric function, while theta is the unknown angle. 
  • Based on this formula, either the unknown angle is determined, the trigonometric equation is simplified or any other trigonometric relation is derived.

Read More: Inverse Trigonometric Functions

Key terms: Sin, Sin Squared X, Triangle, Pythagoras Theorem, Cosine, Secant, Cosecant, Tangent, Cotangent, Perpendicular, Hypotenuse, Integrals, Trigonometric Ratio, Trigonometry


Tangent 3 Theta

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Tan 3 θ can be expressed as a triple angle identity in trigonometry. Tangent 3 Theta is known to be a vital trigonometric identity used to solve a variety of trigonometric and integration issues.

  • For a triple angle, it is a trigonometric function which returns the tan function value.
  • It can also be written as tan 3 θ = sin 3 θ/cos 3 θ because the tangent function is a ratio of the sine functions and cosine functions.
  • The tan 3 θ value repeats once after every π/3 radians, tan3θ = tan (3θ + π/3). In fact, the graph it has is thinner than tan θ’s graph.

Trigonometry is a discipline of mathematics that deals with the use of certain angle functions in computations. In trigonometry, the abbreviations used are sine (sin), cosine (cos), cotangent (cot), tangent (tan), secant (sec), and cosecant (CSC).

Trigonometric Functions Detailed Video Explanation

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Tan 3 Theta Formula

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Tan 3 theta is also known as tan triple angle identity, and it is implemented as a formula in the following two circumstances.

  • The quotient of subtracting Tan cubed of angle from three times Tan of Angle by subtracting three times Tan squared of angle from one is the Tan of Triple angle.
  • Tan of Triple Angle is the quotient of subtracting three times Tan of an Angle from three times Tan of an Angle by subtracting three times Tan squared of an angle from one.

The relationship that can be established between the basic trigonometric functions carrying 3 times the angle in the trigonometric functions of those angles itself typically yields the trigonometric triple-angle identities.

By the triple angle identity, we can expand or simplify the triple angle tan functions like tan 3 A, tan 3 x, tan 3 alpha etc:

Tan 3 theta = 3 tan theta – tan theta / 1 – 3 tantheta

Tan 3 Theta Derivation

Tangent 3 theta formula can be derived by using the sum angle formula used for the Tangent theta and Tangent 2 theta ratios.

In order to show that tan 3θ = (3 tan θ – tan3θ) / (1 – 3 tan2θ), we can express it as: 3θ as (2θ + θ).

Thus,

L.H.S. = tan 3θ 

= tan (2θ + θ)

By applying the formula, tan (x + y) = (tan x + tan y) / (1 – tan x tan y)

= (tan 2θ + tan θ)/(1 – tan 2θ tan θ)

Thus, by using the formula, we get: tan 2x = (2 tan x) / (1 – tan2x) for tan 2θ

= [(2 tan θ / (1 – tan2θ)) + tan θ] / [1 – (2 tan θ / (1 – tan2θ)) tan θ]

= (tan θ – tan3θ + 2 tan θ) / (1 – tan2θ – 2 tan2θ)

= (3 tan θ – tan3θ) / (1 – 3 tan2θ) 

= R.H.S.

Hence, it derives the formula for the tangent 3 theta ratio.

Read More: Introduction to Trigonometry


How to Prove Tan 3 Theta Formula?

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We have,

Tan 3 = (3 tan θ − tan 3 θ) / (1 − 3 tan 2 θ)

Tanking Left-hand side = tan (3 θ) = tan(2 θ + θ)

We must apply the tan(A + B) formula and the tan(A + B) formula to obtain the RHS equation (2A)

As a result, we have:

tan(A + B) = (tanA + tan B) / (1 - tanA tanB) and tan 2A = (2tanA) / (1 - tan2A)

Taking A to be 2θ and B to be θ and use the tan(A + B) formula, we have:

tan( 2θ + θ) =(tan 2 θ + tan θ) / (1 − tan 2 θ × tan θ)

Apply the tan 2 A formula now:

→ \(tan (2 \theta +\theta) = \frac{\frac{2tan\theta}{1-tan \theta} + tan \theta}{1-\frac{2 tan \theta}{1-tan \theta} tan \theta}\)

= 2 tan θ + tan θ

Read More: Heights and Distances


Things to Remember

  • The sign of ratios in different quadrants, involving co-function identities, sum and difference identities, and so on, are some of the trigonometric formulas. 
  • Trigonometry is a discipline of mathematics that deals with the use of certain angle functions in computations. 
  • There are six functions of an angle that are commonly used in trigonometry, namely: sin (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc).
  • The Tangent 3 Theta formula can be expressed as, Tan 3 theta = 3 tan theta – tan theta / 1 – 3 tan
  • The domain, range, and value of a compound trigonometric function are determined using these trigonometric formulas. 

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Previous Year Questions


Sample Questions

Ques. Find the value of x if Tan6A.Tan3A=1. (2 Marks)

Ans. (TanX) . (CotX) = 1 its a formula

Tan6A.Cot6A if equal to 1

Then,

(Tan6A) . (Tan3A) = (Tan6A) . (Cot6A)

Dividing both sides with Tan6A-

Tan3A = Cot6A

Tan(90-3A) = Cot 6A

So, 

3A + 6A = 90

90 / 9 = A

A = 10

Ques. Solve the given equation- tan 3θ = cot θ? (2 Marks)

Ans. Tan 3θ = cot θ = tan (π / 2 − θ)

Comparing the angles, we get-

3θ = n π + (π / 2 − θ)

4θ = (2n + 1) π / 2,

θ = ( 2n + 1 ) π / 8

Ques. Solve the given equation- tan 3θ = − 1 (2 Marks)

Ans. Tan 3θ = − 1 

tan 3θ = tan (−\(\frac{\pi}{4}\))

Comparing angles, we get-

3θ = n π + (− \(\frac{\pi}{4}\)), where n ∈ Z

θ = (\(\frac{n\pi}{4}\)\(\frac{\pi}{4}\)), where n ∈ Z

Ques. Find the value of θ if tan θ + tan 2 θ + \(\sqrt{3}\) tan θ tan 2 θ = \(\sqrt{3}\). (2 Marks)

Ans. As per the given question,

tan θ + tan 2 θ + \(\sqrt{3}\) tan θ tan 2 θ = \(\sqrt{3}\)

\(\frac{tan \theta + tan 2 \theta}{1 - tan \theta tan 2 \theta} \)=\(\sqrt{3}\)

tan 3x = 3

3 θ = n π + 3

\(\theta = \frac{(3n+1) \pi}{9}\)

Ques. What is the value of θ if, tan 3 θ + tan θ = 2 tan 2 θ, where n ∈ Z? (3 marks)

Ans. Tan 3 θ − tan 2 θ = tan 2 θ − tan θ

\(\frac{sin\theta}{cos 3 \theta cos 2 \theta} = \frac{ sin \theta}{cos 2\theta cos \theta}\)

sin θ cos 2θ cos θ = sin θ cos 3θ cos θ

sin θ cos 2θ (cos θ − cos 3θ) = 0

sin θ cos 2θ (2 sin 2θ sin θ) = 0

sin 4θ sin2θ = 0

sin 4θ = 0 

sin2θ = 0

θ = \(\frac{n\pi}{4}\)

θ = n π

Ques. Find the value θ, if tan 3θ − tan 2θ − tan θ = 0? (3 marks)

Ans. 2θ + θ = 3θ 

⇒ tan (2θ+θ) = tan 3θ

\(\frac{tan \theta + tan 2 \theta}{1 - tan \theta tan 2 \theta} \)=tan3

tan 2θ + tanθ = tan 3θ − tan θ tan 2θ tan 3θ

tan 3θ − tan 2θ − tan θ = tan θ tan 2θ tan 3θ

Now,

tan 3θ − tan 2θ − tan θ = 0

tan θ tan 2θ tan 3θ = 0

θ = n π

2 θ = n π 

3 θ = n π

θ = \(\frac{n\pi}{3}\)

Ques. What is the significance of tan 3 theta? (3 Marks)

Ans. The significance of Tan 3 Theta is:

  • The tangent 3 theta formula can be used when simplifying the triple angle tan functions. 
  • It can also be used to expand for further solutions.
  • Tan corresponds to the trigonometric function whereas theta corresponds to the angle a student may have to an unknown angle.

Ques. Solve- [(tan 5θ + tan 3θ) / 4 cos 4θ (tan 5θ - tan 3θ)] (3 Marks)

Ans. [(Tan 5θ + tan 3θ) / 4cos 4θ (tan 5θ - tan 3θ)]

= (sin 5θ cos 3θ + sin 3θ cos 5θ) / [4 cos 4θ sin 5θ cos 3θ - sin 3θ cos 5θ)]

= sin 8θ / (4 cos 4θ sin 2θ)

= 2 sin 4θ cos 4θ / 4 cos 4θ sin 2θ

= 4 sin 2θ cos 2θ / 4 sin 2θ

= cos 2θ

Ques. Solve the given equation- tan 2θ . tan θ = 1? (2 Marks)

Ans. 2 tan θ 1 − tan 2 tan = 1

3 tan 2 θ = 1

tan θ = ± 1/3 = ± tan (π/6)

θ = n π ± π / 6

Ques. Solve- tan 5θ = cot 2θ? (2 Marks)

Ans. Tan 5 θ = cot 2 θ = tan (π/2 − 2 θ)

5θ = nπ + π/2 − 2θ

7θ = nπ + π/2

θ = nπ/7 + π/14, where n ∈ Z

But n = 3, 10, 17... where tan 5θ is not defined.

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CBSE CLASS XII Related Questions

  • 1.
    If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


      • 2.

        Evaluate:
        \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]


          • 3.
            Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


              • 4.
                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 \( \cos^{-1}(x^2) \).


                  • 5.
                    Find:

                    The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

                      • \(-\frac{\pi}{2}\)
                      • \(-\frac{\pi}{4}\)
                      • \(\frac{\pi}{4}\)
                      • \(\frac{\pi}{2}\)

                    • 6.
                      Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

                        CBSE CLASS XII Previous Year Papers

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