Question

If $x = $\[\tan \dfrac{\pi }{{18}}\] then \[3x_{}^6 - 27x_{}^4 + 33x_{}^2\] equals to-
A. \[1\]
B. $2$
C. $3\sqrt 3 $
D. $\dfrac{1}{3}$

Answer 1

Hint: First we can solve the question by applying the formula.
We need to substitute the given value of $x$ in the formula
Finally we get the required solution.

Formula used: $\tan 3\theta = \dfrac{{3\tan \theta - \tan _{}^3\theta }}{{1 - 3\tan _{}^2\theta }}$

Complete step-by-step answer:
It is given that the value \[x = \tan \dfrac{\pi }{{18}}\]
Let us considered $\theta = x$
Substitute the above in the formula and we can write $\tan 3x = \dfrac{{3\tan x - \tan _{}^3x}}{{1 - 3\tan _{}^2x}}....\left( 1 \right)$
Putting the value of $x = \tan \left( {\dfrac{\pi }{{18}}} \right)$ in equation \[\left( 1 \right)\] we get
$\tan \left( {3.\dfrac{\pi }{{18}}} \right) = \dfrac{{3\tan \left( {\dfrac{\pi }{{18}}} \right) - \tan _{}^3\left( {\dfrac{\pi }{{18}}} \right)}}{{1 - 3\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right)}}$
Now in left side, if we look we can simplify, $\tan \left( {\dfrac{{3\pi }}{{18}}} \right) = \tan \left( {\dfrac{\pi }{6}} \right)$
$\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{{3\tan \left( {\dfrac{\pi }{{18}}} \right) - \tan _{}^3\left( {\dfrac{\pi }{{18}}} \right)}}{{1 - 3\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right)}}$
Here $\pi = {180^ \circ }$ and we can simplify it,
Putting the value of $\tan \left( {\dfrac{\pi }{6}} \right)$$ = \tan {30^ \circ }$$ = \dfrac{1}{{\sqrt 3 }}$ in the equation we get-
$\dfrac{1}{{\sqrt 3 }} = \dfrac{{3\tan \left( {\dfrac{\pi }{{18}}} \right) - \tan _{}^3\left( {\dfrac{\pi }{{18}}} \right)}}{{1 - 3\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right)}}$
Now, we can squaring on both the sides we get
$\left( {\dfrac{1}{{\sqrt 3 }}} \right)_{}^2 = \left[ {\dfrac{{3\tan \left( {\dfrac{\pi }{{18}}} \right) - \tan _{}^3\left( {\dfrac{\pi }{{18}}} \right)}}{{1 - 3\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right)}}} \right]_{}^2$
After squaring both the sides we get $\dfrac{1}{3}$ on the left side
Also we need to apply the formula of $(a - b)_{}^2 = a_{}^2 + b_{}^2 - 2ab$ on both the numerator and denominator of the right side we can write it as,
$\dfrac{1}{3} = \dfrac{{9\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right) + \tan _{}^6\left( {\dfrac{\pi }{{18}}} \right) - 6\tan _{}^4\left( {\dfrac{\pi }{{18}}} \right)}}{{1 + 9\tan _{}^4\left( {\dfrac{\pi }{{18}}} \right) - 6\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right)}}$
Now we can take the above step in to cross multiplication on both the sides for getting the easy simplification
 $1 + 9\tan _{}^4\left( {\dfrac{\pi }{{18}}} \right) - 6\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right) = 27\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right) + 3\tan _{}^6\left( {\dfrac{\pi }{{18}}} \right) - 18\tan _{}^4\left( {\dfrac{\pi }{{18}}} \right)$
We can add and subtracting the above terms in the form of equating itself, and \[1\] take it as RHS
$3\tan _{}^6\left( {\dfrac{\pi }{{18}}} \right) - 27\tan _{}^4\left( {\dfrac{\pi }{{18}}} \right) + 33\tan _{}^2\left( {\dfrac{\pi }{{18}}} \right) = 1$
As mentioned above we have to take the given value \[x = \tan \dfrac{\pi }{{18}}\],
So we can write it as,
Thus the value of \[3x_{}^6 - 27x_{}^4 + 33x_{}^2\] is \[1\]

Note: There are two ways of solving this question. One is by applying the formula of $\tan 3\theta = \dfrac{{3\tan \theta - \tan _{}^3\theta }}{{1 - 3\tan _{}^2\theta }}$ and another is finding out the value of $\tan \dfrac{\pi }{{18}}$ and substituting it’s value in \[3x_{}^6 - 27x_{}^4 + 33x_{}^2\].
Both are correct but you should choose one which is suitable for you.
But it is an easy way to solve this question.
It must be kept in mind for solving the question of trigonometry you must know all the formulas related to it.