Question

Find the value of \[\dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \]
A. $\dfrac{tan 3\theta}{tan \theta}$
B. $\dfrac{cot 3\theta}{cot \theta}$
C. \[\tan 3\theta \tan \theta \]
D. \[\cot 3\theta \cot \theta \]

Answer 1

Hint: We use the formula of \[{a^2} - {b^2} = (a - b)(a + b)\] to write both numerator and denominator in simple form. Shuffle and pair the elements of numerator and denominator to form pairs that combine and give trigonometric identities of \[\tan (A + B)\] and \[\tan (A - B)\]. Add and subtract the angles within to get the answer.
* Formula of \[\tan (A + B)\]is given as \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
* Formula of \[\tan (A - B)\]is given as \[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]

Complete step by step solution:
We have to find the value of \[\dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }}\]............… (1)
Since we know \[{a^2} - {b^2} = (a - b)(a + b)\]
We have a numerator as \[{\tan ^2}2\theta - {\tan ^2}\theta \].
On comparing with the formula \[{a^2} - {b^2} = (a - b)(a + b)\], \[a = \tan 2\theta ,b = \tan \theta \]
\[ \Rightarrow {\tan ^2}2\theta - {\tan ^2}\theta = (\tan 2\theta - \tan \theta )(\tan 2\theta + \tan \theta )\]............… (2)
We have a denominator as \[1 - {\tan ^2}2\theta {\tan ^2}\theta \].
On comparing with the formula \[{a^2} - {b^2} = (a - b)(a + b)\], \[a = 1,b = \tan 2\theta \tan \theta \]
\[ \Rightarrow 1 - {\tan ^2}2\theta {\tan ^2}\theta = (1 - \tan 2\theta \tan \theta )(1 + \tan 2\theta \tan \theta )\]...............… (3)
Substitute the values of numerator from equation (2)and value of denominator from equation (3) in equation (1)
\[ \Rightarrow \dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \dfrac{{(\tan 2\theta - \tan \theta )(\tan 2\theta + \tan \theta )}}{{(1 - \tan 2\theta \tan \theta )(1 + \tan 2\theta \tan \theta )}}\]................… (4)
Now we know the trigonometric identities \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]and\[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\].
We pair the terms in RHS of equation (4) in such a way that the given trigonometric identities can be used to form a simpler solution.
\[ \Rightarrow \dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \left\{ {\dfrac{{(\tan 2\theta - \tan \theta )}}{{(1 + \tan 2\theta \tan \theta )}}} \right\}\left\{ {\dfrac{{(\tan 2\theta + \tan \theta )}}{{(1 - \tan 2\theta \tan \theta )}}} \right\}\]...............… (5)
We can see the first bracket is similar to the identity\[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]and the second bracket is similar to the identity\[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\].
We write the simpler form of fractions using the identities.
\[ \Rightarrow \dfrac{{(\tan 2\theta - \tan \theta )}}{{(1 + \tan 2\theta \tan \theta )}} = \tan (2\theta - \theta )\] and \[\dfrac{{(\tan 2\theta + \tan \theta )}}{{(1 - \tan 2\theta \tan \theta )}} = \tan (2\theta + \theta )\]
Substitute the values back in the equation (5)
\[ \Rightarrow \dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \left\{ {\tan (2\theta - \theta )} \right\}\left\{ {\tan (2\theta + \theta )} \right\}\]
Add and subtract the angles as required in the bracket.
 \[ \Rightarrow \dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \left\{ {\tan \theta } \right\}\left\{ {\tan 3\theta } \right\}\]
\[ \Rightarrow \dfrac{{{{\tan }^2}2\theta - {{\tan }^2}\theta }}{{1 - {{\tan }^2}2\theta {{\tan }^2}\theta }} = \tan 3\theta \tan \theta \]
Since RHS is \[\tan 3\theta \tan \theta \]

\[\therefore \]Correct option is C.

Note: Students many times make the mistake of solving the question by general multiplication after they open up the values using \[{a^2} - {b^2} = (a - b)(a + b)\]. This process will be very long and will have complex calculations; first we will multiply all the values then cancel terms and then take common factors.
Students many times try to solve the equation by substituting the value for \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\tan 2\theta = \dfrac{{\sin 2\theta }}{{\cos 2\theta }}\], which will make the solution very complex and students should avoid this process of solving.