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Question: Write the function in the simplest form: \({\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - ...

Write the function in the simplest form: tan⁡−1(3a2x−x3a3−3ax2),a>0;−a3⩽x⩽a3{\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 }}

Explanation

Solution

We will make a suitable substitution for xx in terms of ‘tan⁡\tan ’ in the given function. Then we will simplify the given function in terms of ‘tan⁡\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⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}
  2. tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}(\tan \theta ) = \theta

Complete step by step solution:
The function given is
tan⁡−1(3a2x−x3a3−3ax2),a>0;−a3⩽x⩽a3{\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 }}………………………………………..(1)\left( 1 \right)
To simplify the function, we will make a substitution for xx in terms of the function ‘tan⁡\tan ’ . We make this substitution in order to eliminate the variables aa and xx.
Let us substitute x=atan⁡θx = a\tan \theta .
Taking aa to the other side, we get
xa=tan⁡θ\dfrac{x}{a} = \tan \theta .
Applying tan⁡−1{\tan ^{ - 1}}on both sides, we get
tan⁡−1(xa)=tan⁡−1(tan⁡θ){\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) = {\tan ^{ - 1}}(\tan \theta ).
Using the property tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}(\tan \theta ) = \theta , we have
tan⁡−1(xa)=θ{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) = \theta …………………………………….(2)\left( 2 \right)
Substituting x=atan⁡θx = a\tan \theta in the equation (1)\left( 1 \right), we get tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(3a2(atan⁡θ)−(atan⁡θ)3a3−3a(atan⁡θ)2){\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,
⇒tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(3a3tan⁡θ−a3tan⁡3θa3−3a3tan⁡2θ)\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 a3{a^3} common outside from the numerator and denominator on the RHS. Therefore, we get
⇒tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(a3(3tan⁡θ−tan⁡3θ)a3(1−3tan⁡2θ))\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 a3{a^3} from the numerator and denominator, we get
⇒tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)\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⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}. Thus, we get
tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(tan⁡3θ){\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\tan 3\theta } \right)……………………………(3)\left( 3 \right)
We will use the property tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}(\tan \theta ) = \theta in equation (3)\left( 3 \right). Therefore,
tan⁡−1(3a2x−x3a3−3ax2)=tan⁡−1(tan⁡3θ)=3θ{\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 ……………………………(4)\left( 4 \right)
Now, we will use equation (2)\left( 2 \right) in equation (4)\left( 4 \right) and finally, we get the simplest form the function as tan⁡−1(3a2x−x3a3−3ax2)=3tan⁡−1(xa){\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=atan⁡θx = a\tan \theta , where aa is a constant. If the function wastan⁡−1(3x−x31−3x2){\tan ^{ - 1}}\left( {\dfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), where a=1a = 1, then the substitution would simply be x=tan⁡θx = \tan \theta . We have used a substitution in terms of ′tan⁡′'\tan ' since the outer function is tan⁡−1{\tan ^{ - 1}} also tan⁡\tan and tan⁡−1{\tan ^{ - 1}} are inverse functions.
We use trigonometric identities only in equations where the trigonometric function is present.