# Find the general value of $\theta $, if $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$.

(A) $\left[ {n\pi + \dfrac{\pi }{5}} \right]$ (B) $\left( {n + \dfrac{1}{6}} \right)\dfrac{\pi }{5}$ (C) $\left( {2n \pm \dfrac{1}{6}} \right)\dfrac{\pi }{6}$ (D) $\left( {n + \dfrac{1}{3}} \right)\dfrac{\pi }{5}$

Last updated date: 24th Mar 2023

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Answer

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Hint: Convert the given equation in equation containing $\tan 5\theta $ using the formula:

$ \Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

Complete step-by-step answer:

The given equation is $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$. This can be written as:

\[

\Rightarrow \sqrt 3 \left( {\tan 2\theta + \tan 3\theta } \right) + \tan 2\theta \tan 3\theta = 1, \\

\Rightarrow \sqrt 3 \left( {\tan 2\theta + \tan 3\theta } \right) = 1 - \tan 2\theta \tan 3\theta , \\

\Rightarrow \sqrt 3 \dfrac{{\left( {\tan 2\theta + \tan 3\theta } \right)}}{{1 - \tan 2\theta \tan 3\theta }} = 1, \\

\Rightarrow \dfrac{{\tan 2\theta + \tan 3\theta }}{{1 - \tan 2\theta \tan 3\theta }} = \dfrac{1}{{\sqrt 3 }} ....(i) \\

\]

We know that, $\dfrac{1}{{\sqrt 3 }} = \tan \dfrac{\pi }{6}$ and $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$, if we substitute $A = 2\theta $ and $B = 3\theta $, we have $\dfrac{{\tan 2\theta + \tan 3\theta }}{{1 - \tan 2\theta \tan 3\theta }} = \tan 5\theta $. Putting these values in equation \[(i)\], we’ll get:

$

\Rightarrow \tan 5\theta = \tan \dfrac{\pi }{6}, \\

\Rightarrow 5\theta = n\pi + \dfrac{\pi }{6}, \\

\Rightarrow \theta = \dfrac{1}{5}\left( {n\pi + \dfrac{\pi }{6}} \right), \\

\Rightarrow \theta = \left( {n + \dfrac{1}{6}} \right)\dfrac{\pi }{5} \\

$

Thus, the general value of $\theta $ is $\left( {n + \dfrac{1}{6}} \right)\dfrac{\pi }{5}$. (B) is the correct option.

Note: In the above equation, we get $\tan 5\theta = \tan \dfrac{\pi }{6}$. As we know that $\tan x$ is periodic with period $\pi $, that’s why instead of $5\theta = \dfrac{\pi }{6}$ we have to take $5\theta = n\pi + \dfrac{\pi }{6}$. From this we can say that the value of $5\theta $ will repeat after every integral multiple of $\pi $.

$ \Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

Complete step-by-step answer:

The given equation is $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$. This can be written as:

\[

\Rightarrow \sqrt 3 \left( {\tan 2\theta + \tan 3\theta } \right) + \tan 2\theta \tan 3\theta = 1, \\

\Rightarrow \sqrt 3 \left( {\tan 2\theta + \tan 3\theta } \right) = 1 - \tan 2\theta \tan 3\theta , \\

\Rightarrow \sqrt 3 \dfrac{{\left( {\tan 2\theta + \tan 3\theta } \right)}}{{1 - \tan 2\theta \tan 3\theta }} = 1, \\

\Rightarrow \dfrac{{\tan 2\theta + \tan 3\theta }}{{1 - \tan 2\theta \tan 3\theta }} = \dfrac{1}{{\sqrt 3 }} ....(i) \\

\]

We know that, $\dfrac{1}{{\sqrt 3 }} = \tan \dfrac{\pi }{6}$ and $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$, if we substitute $A = 2\theta $ and $B = 3\theta $, we have $\dfrac{{\tan 2\theta + \tan 3\theta }}{{1 - \tan 2\theta \tan 3\theta }} = \tan 5\theta $. Putting these values in equation \[(i)\], we’ll get:

$

\Rightarrow \tan 5\theta = \tan \dfrac{\pi }{6}, \\

\Rightarrow 5\theta = n\pi + \dfrac{\pi }{6}, \\

\Rightarrow \theta = \dfrac{1}{5}\left( {n\pi + \dfrac{\pi }{6}} \right), \\

\Rightarrow \theta = \left( {n + \dfrac{1}{6}} \right)\dfrac{\pi }{5} \\

$

Thus, the general value of $\theta $ is $\left( {n + \dfrac{1}{6}} \right)\dfrac{\pi }{5}$. (B) is the correct option.

Note: In the above equation, we get $\tan 5\theta = \tan \dfrac{\pi }{6}$. As we know that $\tan x$ is periodic with period $\pi $, that’s why instead of $5\theta = \dfrac{\pi }{6}$ we have to take $5\theta = n\pi + \dfrac{\pi }{6}$. From this we can say that the value of $5\theta $ will repeat after every integral multiple of $\pi $.

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