Question

Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$

Answer 1

Hint:
We will make a suitable substitution for $x$ in terms of ‘$\tan $’ in the given function. Then we will simplify the given function in terms of ‘$\tan $’ in order to apply an appropriate identity. This will give us the simplest form of the function.

Formula used:
We will use the following formulas:
1) $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$
2) ${\tan ^{ - 1}}(\tan \theta ) = \theta $

Complete step by step solution:
The function given is
 ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$………………………………………..$\left( 1 \right)$
To simplify the function, we will make a substitution for $x$ in terms of the function ‘$\tan $’ . We make this substitution in order to eliminate the variables $a$ and $x$.
Let us substitute $x = a\tan \theta $ .
Taking $a$ to the other side, we get
$\dfrac{x}{a} = \tan \theta $.
 Applying ${\tan ^{ - 1}}$on both sides, we get
${\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) = {\tan ^{ - 1}}(\tan \theta )$.
Using the property ${\tan ^{ - 1}}(\tan \theta ) = \theta $, we have
${\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) = \theta $…………………………………….$\left( 2 \right)$
Substituting $x = a\tan \theta $ in the equation $\left( 1 \right)$, we get ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{3{a^2}(a\tan \theta ) - {{(a\tan \theta )}^3}}}{{{a^3} - 3a{{(a\tan \theta )}^2}}}} \right)$
Expanding the terms in the numerator and denominator on the RHS,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{3{a^3}\tan \theta - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)\]
Let us take ${a^3}$ common outside from the numerator and denominator on the RHS. Therefore, we get
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{{a^3}(3\tan \theta - {{\tan }^3}\theta )}}{{{a^3}(1 - 3{{\tan }^2}\theta )}}} \right)\]
Cancelling out ${a^3}$ from the numerator and denominator, we get
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)\]
Now, on the RHS, we will use the identity $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$. Thus, we get
\[{\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\tan 3\theta } \right)\]……………………………$\left( 3 \right)$
We will use the property ${\tan ^{ - 1}}(\tan \theta ) = \theta $ in equation $\left( 3 \right)$. Therefore,
\[{\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\tan 3\theta } \right) = 3\theta \] ……………………………$\left( 4 \right)$
Now, we will use equation $\left( 2 \right)$ in equation $\left( 4 \right)$ and finally, we get the simplest form the function as \[{\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = 3{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)\]

Note:
We have used a substitution $x = a\tan \theta $, where $a$ is a constant. If the function was\[{\tan ^{ - 1}}\left( {\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)\], where $a = 1$, then the substitution would simply be $x = \tan \theta $. We have used a substitution in terms of $'\tan '$ since the outer function is ${\tan ^{ - 1}}$ also $\tan $ and ${\tan ^{ - 1}}$ are inverse functions.
We use trigonometric identities only in equations where the trigonometric function is present.