Question

For any $a,b,x,y > 0$, prove that if: $\dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3a{b^2} - {a^3}}}{{{b^3} - 3{a^2}b}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3x{y^2} - {x^3}}}{{{y^3} - 3{x^2}y}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$, then $\alpha = - ax + by,\beta = bx + ay$.

Answer 1

Here, we will simplify the equation and then solve it by using the basic trigonometric function. Then by comparing the left hand side and right hand side of the equation we will get the value of $\alpha ,\beta$.

Complete step by step solution:
Given equation is $\dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3a{b^2} - {a^3}}}{{{b^3} - 3{a^2}b}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3x{y^2} - {x^3}}}{{{y^3} - 3{x^2}y}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$.
First, we will divide the numerator and denominator of the first term of the LHS of the equation by ${b^3}$. Therefore, we get
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{3a{b^2} - {a^3}}}{{{b^3}}}}}{{\dfrac{{{b^3} - 3{a^2}b}}{{{b^3}}}}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3x{y^2} - {x^3}}}{{{y^3} - 3{x^2}y}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\dfrac{a}{b} - \dfrac{{{a^3}}}{{{b^3}}}}}{{1 - 3\dfrac{{{a^2}}}{{{b^2}}}}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3x{y^2} - {x^3}}}{{{y^3} - 3{x^2}y}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we will divide the numerator and denominator of the second term of the LHS of the equation by ${y^3}$. Therefore, we get
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\dfrac{a}{b} - \dfrac{{{a^3}}}{{{b^3}}}}}{{1 - 3\dfrac{{{a^2}}}{{{b^2}}}}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{3x{y^2} - {x^3}}}{{{y^3}}}}}{{\dfrac{{{y^3} - 3{x^2}y}}{{{y^3}}}}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\dfrac{a}{b} - \dfrac{{{a^3}}}{{{b^3}}}}}{{1 - 3\dfrac{{{a^2}}}{{{b^2}}}}}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\dfrac{x}{y} - \dfrac{{{x^3}}}{{{y^3}}}}}{{1 - 3\dfrac{{{x^2}}}{{{y^2}}}}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we will let $\dfrac{a}{b} = \tan \theta$ and $\dfrac{x}{y} = \tan \phi$ also we get $\theta = {\tan ^{ - 1}}\dfrac{a}{b}$ and $\phi = {\tan ^{ - 1}}\dfrac{x}{y}$. Therefore, by putting this value in the above equation we get
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\tan \phi - {{\tan }^3}\phi }}{{1 - 3{{\tan }^2}\phi }}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
We know that the value of the $\tan 3\theta$ is $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$. Therefore, we get
$\Rightarrow \dfrac{2}{3}{\tan ^{ - 1}}\left( {\tan 3\theta } \right) + \dfrac{2}{3}{\tan ^{ - 1}}\left( {\tan 3\phi } \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now the inverse function term will cancel out with the normal function, we get
$\Rightarrow \dfrac{2}{3} \times 3\theta + \dfrac{2}{3} \times 3\phi = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
$\Rightarrow 2\theta + 2\phi = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we will put the value of $\theta ,\phi$ in the above equation, we get
$\Rightarrow 2{\tan ^{ - 1}}\dfrac{a}{b} + 2{\tan ^{ - 1}}\dfrac{x}{y} = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
$\Rightarrow 2\left( {{{\tan }^{ - 1}}\dfrac{a}{b} + {{\tan }^{ - 1}}\dfrac{x}{y}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
We know the formula of the ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B$ which is equals to ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\dfrac{{A + B}}{{1 - AB}}$, we get
$\Rightarrow 2\left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{a}{b} + \dfrac{x}{y}}}{{1 - \dfrac{a}{b}\dfrac{x}{y}}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we will solve and simplify this equation, we get
$\Rightarrow 2\left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{{ay + bx}}{{by}}}}{{\dfrac{{by - ax}}{{by}}}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
$\Rightarrow 2\left( {{{\tan }^{ - 1}}\dfrac{{ay + bx}}{{by - ax}}} \right) = {\tan ^{ - 1}}\dfrac{{2\alpha \beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we will simplify the RHS of the equation. We will divide the numerator and denominator of the RHS of the equation by ${\alpha ^2}$, we get
$\Rightarrow 2\left( {{{\tan }^{ - 1}}\dfrac{{ay + bx}}{{by - ax}}} \right) = {\tan ^{ - 1}}\dfrac{{\dfrac{{2\alpha \beta }}{{{\alpha ^2}}}}}{{\dfrac{{{\alpha ^2} - {\beta ^2}}}{{{\alpha ^2}}}}} = {\tan ^{ - 1}}\dfrac{{2\dfrac{\beta }{\alpha }}}{{1 - \dfrac{{{\beta ^2}}}{{{\alpha ^2}}}}}$
We know that the formula of $2{\tan ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{{2A}}{{1 - {A^2}}}$. Therefore, by comparing the RHS of the equation with this formula we get
$\Rightarrow 2{\tan ^{ - 1}}\dfrac{{ay + bx}}{{by - ax}} = 2{\tan ^{ - 1}}\dfrac{\beta }{\alpha }$
Now by simply comparing the LHS and RHS of the equation we get the value of $\alpha ,\beta$, we get
$\alpha = - ax + by$ and $\beta = bx + ay$
Hence, proved.

Note:
For solving this question, we need to know different trigonometric identities and formulas. If we know the trigonometric identities then we will be able to understand what we need to do such that the given expression is converted into one of the identities. Trigonometric identities are derived from six basic trigonometric ratios and can be only used in an equation or expression where trigonometric function is involved.