
If \[\dfrac{{\tan 3\theta - 1}}{{\tan 3\theta + 1}} = \sqrt 3 \], then the general value of \[\theta \]is
A. \[\dfrac{{n\pi }}{3} + \dfrac{\pi }{{12}}\]
B. \[\dfrac{{n\pi }}{3} + \dfrac{{7\pi }}{{36}}\]
C. \[n\pi + \dfrac{{7\pi }}{{12}}\]
D. \[n\pi + \dfrac{\pi }{{12}}\]
Answer
162.6k+ views
Hint: We use the Componendo-Dividendo rule, also known as Componendo and Dividend, to compare and analyse ratios and proportions. It is notably helpful for solving fractional equations or rational equations in mathematical Olympiads, especially when fractions are present. To derive the equation, we use trigonometry formulas. The nature of all trigonometric identities is cyclicity. After this periodicity is constant, they repeat. Formulas involving trigonometric functions are known as trigonometric idents. For each possible value of the variables, these identities hold true. The relationship between the length of the sides of the right triangle and the measurement of the angles is known as the trigonometric ratio.
Formula Used: By using Componendo-Dividendo rule,
It can be expressed as \[a:b = c:d\] and \[\left( {a + b} \right):\left( {a - b} \right) = \left( {c + d} \right):(c - d)\]
\[\tan \dfrac{\pi }{3} = \sqrt 3 \]and
\[\tan \dfrac{\pi }{4} = 1\]
Complete step by step solution:
\[\dfrac{{\left[ {\tan 3\theta - 1 + \tan 3\theta + 1} \right]}}{{\left[ {\tan 3\theta - 1 - \tan 3\theta - 1} \right]}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\]
\[\dfrac{{2\tan 3\theta }}{{ - 2}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\]
\[\tan 3\theta = \dfrac{{\sqrt 3 + 1}}{{1 - \sqrt 3 }}\]
Simplify on both sides
\[\dfrac{{\tan 3\theta - 1}}{{\tan 3\theta + 1}} = \sqrt 3 \]
\[\dfrac{{\tan 3\theta - \tan (\dfrac{\pi }{4})}}{{1 + \tan 3\theta .\tan (\dfrac{\pi }{4})}} = \sqrt 3 \]
From the formula \[\tan \dfrac{\pi }{3} = \sqrt 3 \] and \[\tan \dfrac{\pi }{4} = 1\]
\[\tan (3\theta - \dfrac{\pi }{4}) = \tan \dfrac{\pi }{3}\]
Cancel \[\tan \] on both sides
\[3\theta - (\dfrac{\pi }{4}) = n\pi + (\dfrac{\pi }{3})\]
\[3\theta = n\pi + \dfrac{\pi }{{12}}\]
Divide by \[3\] on both sides
\[\theta = \dfrac{{n\pi }}{3} + \dfrac{{7\pi }}{{36}}\]
The general value of \[\theta = \dfrac{{n\pi }}{3} + \dfrac{{7\pi }}{{36}}\]
Option ‘B’ is correct
Note: It is important to note the Componendo-Dividendo rule, Therefore, the ratios of the sum of the numerator and denominator to the difference of the numerator and denominator of both rational numbers are equal if the ratio of any two numbers is equal to the ratio of another two numbers. If one quantity increases while the other drops by the same amount, then the two quantities are said to be indirectly proportional. If the rise (or decrease) in one quantity causes an equal increase (or decrease) in the other, the two quantities are said to be directly proportional. For various trigonometric identities, this periodicity constant varies.
Formula Used: By using Componendo-Dividendo rule,
It can be expressed as \[a:b = c:d\] and \[\left( {a + b} \right):\left( {a - b} \right) = \left( {c + d} \right):(c - d)\]
\[\tan \dfrac{\pi }{3} = \sqrt 3 \]and
\[\tan \dfrac{\pi }{4} = 1\]
Complete step by step solution:
\[\dfrac{{\left[ {\tan 3\theta - 1 + \tan 3\theta + 1} \right]}}{{\left[ {\tan 3\theta - 1 - \tan 3\theta - 1} \right]}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\]
\[\dfrac{{2\tan 3\theta }}{{ - 2}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\]
\[\tan 3\theta = \dfrac{{\sqrt 3 + 1}}{{1 - \sqrt 3 }}\]
Simplify on both sides
\[\dfrac{{\tan 3\theta - 1}}{{\tan 3\theta + 1}} = \sqrt 3 \]
\[\dfrac{{\tan 3\theta - \tan (\dfrac{\pi }{4})}}{{1 + \tan 3\theta .\tan (\dfrac{\pi }{4})}} = \sqrt 3 \]
From the formula \[\tan \dfrac{\pi }{3} = \sqrt 3 \] and \[\tan \dfrac{\pi }{4} = 1\]
\[\tan (3\theta - \dfrac{\pi }{4}) = \tan \dfrac{\pi }{3}\]
Cancel \[\tan \] on both sides
\[3\theta - (\dfrac{\pi }{4}) = n\pi + (\dfrac{\pi }{3})\]
\[3\theta = n\pi + \dfrac{\pi }{{12}}\]
Divide by \[3\] on both sides
\[\theta = \dfrac{{n\pi }}{3} + \dfrac{{7\pi }}{{36}}\]
The general value of \[\theta = \dfrac{{n\pi }}{3} + \dfrac{{7\pi }}{{36}}\]
Option ‘B’ is correct
Note: It is important to note the Componendo-Dividendo rule, Therefore, the ratios of the sum of the numerator and denominator to the difference of the numerator and denominator of both rational numbers are equal if the ratio of any two numbers is equal to the ratio of another two numbers. If one quantity increases while the other drops by the same amount, then the two quantities are said to be indirectly proportional. If the rise (or decrease) in one quantity causes an equal increase (or decrease) in the other, the two quantities are said to be directly proportional. For various trigonometric identities, this periodicity constant varies.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
