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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
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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.