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

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.