A four degree polynomial equation in x is given in which if ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ are roots of the equation ${{x}^{4}}-{{x}^{3}}\sin 2\beta +{{x}^{2}}\cos 2\beta -x\cos \beta -\sin \beta =0$ then find the value of:
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$
Answer
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Hint: Using the given equation of degree four, we can write the roots ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ as sum of the roots, sum of the roots taken two at a time, sum of the roots taken three at a time and product of the roots then use these expressions in the expansion of ${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$. You can expand this expression by taking two expressions at a time $\left( {{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}} \right)$ and then use the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$.
Complete step-by-step answer:
The polynomial equation given in the question is:
${{x}^{4}}-{{x}^{3}}\sin 2\beta +{{x}^{2}}\cos 2\beta -x\cos \beta -\sin \beta =0$
The roots of the above equation is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$.
We know that using this polynomial equation of degree 4:
We can write the sum of the roots as follows:
$\begin{align}
& \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=-\dfrac{-\sin 2\beta }{1} \right) \\
& \Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=\sin 2\beta ..............Eq.(1) \\
\end{align}$
We can write sum of the roots taken two at a time as follows:
${{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{4}}+{{x}_{4}}{{x}_{1}}+{{x}_{1}}{{x}_{3}}+{{x}_{2}}{{x}_{4}}=\cos 2\beta $……………. Eq. (2)
We can write sum of the roots taken three at a time as follows:
${{x}_{1}}{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{3}}{{x}_{4}}+{{x}_{3}}{{x}_{4}}{{x}_{1}}+{{x}_{4}}{{x}_{1}}{{x}_{2}}=\cos \beta $…………. Eq. (3)
We can write product of the roots as follows:
${{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=-\sin \beta $……….. Eq. (4)
We have to find the value of:
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$
We are going to find the value of the above expression by taking the sum of first two terms and the sum of last two terms using the following formula:
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)$…………. Eq. (5)
${{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}={{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)$………….. Eq. (6)
Now, adding eq. (5) and eq. (6) we get,
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)$
Solving the right hand side of the above equation we get,
\[{{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)\]
Applying the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$ in the above expression we get,
\[{{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)}+\dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)}}{1-\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)} \right)\]
Solving the R.H.S of the above equation we get,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{\left( {{x}_{1}}+{{x}_{2}} \right)\left( 1-{{x}_{3}}{{x}_{4}} \right)+\left( {{x}_{3}}+{{x}_{4}} \right)\left( 1-{{x}_{1}}{{x}_{2}} \right)}{\left( 1-{{x}_{1}}{{x}_{2}} \right)\left( 1-{{x}_{3}}{{x}_{4}} \right)-\left( {{x}_{1}}+{{x}_{2}} \right)\left( {{x}_{3}}+{{x}_{4}} \right)} \right) \\
& ={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}-\left( {{x}_{1}}{{x}_{3}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}{{x}_{4}} \right)+{{x}_{3}}+{{x}_{4}}-\left( {{x}_{3}}{{x}_{1}}{{x}_{2}}+{{x}_{4}}{{x}_{1}}{{x}_{2}} \right)}{1-\left( {{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}} \right)+{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}-\left( {{x}_{1}}{{x}_{3}}+{{x}_{1}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{4}} \right)} \right) \\
& ={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}-\left( {{x}_{1}}{{x}_{3}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}{{x}_{4}}+{{x}_{3}}{{x}_{1}}{{x}_{2}}+{{x}_{4}}{{x}_{1}}{{x}_{2}} \right)}{1+{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}-\left( {{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}}+{{x}_{1}}{{x}_{3}}+{{x}_{1}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{4}} \right)} \right) \\
\end{align}$
In the above expression substitute the values of sum of the roots, sum of the roots taken two at a time, sum of the roots taken three at a time and product of the roots from $eq.\left( 1,2,3,4 \right)$.
${{\tan }^{-1}}\left( \dfrac{\sin 2\beta -\cos \beta }{1-\sin \beta -\cos 2\beta } \right)$
In the above expression, we can write:
$\begin{align}
& \sin 2\beta =2\sin \beta \cos \beta \\
& 1-\cos 2\beta =2{{\sin }^{2}}\beta \\
\end{align}$
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{2\sin \beta \cos \beta -\cos \beta }{2{{\sin }^{2}}\beta -\sin \beta } \right) \\
& ={{\tan }^{-1}}\left( \dfrac{\cos \beta \left( 2\sin \beta -1 \right)}{\sin \beta \left( 2\sin \beta -1 \right)} \right) \\
\end{align}$
In the above expression $2\sin \beta -1$ will be cancelled out from the numerator and the denominator.
${{\tan }^{-1}}\left( \cot \beta \right)$
From the above solution, we have got the value of ${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$ as ${{\tan }^{-1}}\left( \cot \beta \right)$.
Note: Be careful in writing the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$. You might wrongly write the signs in this formula like in the numerator you can see a plus sign so instead of writing a plus sign, you could write minus sign and in the denominator, you can see a minus sign so instead of writing minus sign you can put plus sign.
Blunders in putting signs are very commonly observed.
Complete step-by-step answer:
The polynomial equation given in the question is:
${{x}^{4}}-{{x}^{3}}\sin 2\beta +{{x}^{2}}\cos 2\beta -x\cos \beta -\sin \beta =0$
The roots of the above equation is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$.
We know that using this polynomial equation of degree 4:
We can write the sum of the roots as follows:
$\begin{align}
& \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=-\dfrac{-\sin 2\beta }{1} \right) \\
& \Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=\sin 2\beta ..............Eq.(1) \\
\end{align}$
We can write sum of the roots taken two at a time as follows:
${{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{4}}+{{x}_{4}}{{x}_{1}}+{{x}_{1}}{{x}_{3}}+{{x}_{2}}{{x}_{4}}=\cos 2\beta $……………. Eq. (2)
We can write sum of the roots taken three at a time as follows:
${{x}_{1}}{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{3}}{{x}_{4}}+{{x}_{3}}{{x}_{4}}{{x}_{1}}+{{x}_{4}}{{x}_{1}}{{x}_{2}}=\cos \beta $…………. Eq. (3)
We can write product of the roots as follows:
${{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=-\sin \beta $……….. Eq. (4)
We have to find the value of:
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$
We are going to find the value of the above expression by taking the sum of first two terms and the sum of last two terms using the following formula:
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)$…………. Eq. (5)
${{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}={{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)$………….. Eq. (6)
Now, adding eq. (5) and eq. (6) we get,
${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)$
Solving the right hand side of the above equation we get,
\[{{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)\]
Applying the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$ in the above expression we get,
\[{{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)}+\dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)}}{1-\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{1-{{x}_{1}}\left( {{x}_{2}} \right)} \right)\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{1-{{x}_{3}}\left( {{x}_{4}} \right)} \right)} \right)\]
Solving the R.H.S of the above equation we get,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{\left( {{x}_{1}}+{{x}_{2}} \right)\left( 1-{{x}_{3}}{{x}_{4}} \right)+\left( {{x}_{3}}+{{x}_{4}} \right)\left( 1-{{x}_{1}}{{x}_{2}} \right)}{\left( 1-{{x}_{1}}{{x}_{2}} \right)\left( 1-{{x}_{3}}{{x}_{4}} \right)-\left( {{x}_{1}}+{{x}_{2}} \right)\left( {{x}_{3}}+{{x}_{4}} \right)} \right) \\
& ={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}-\left( {{x}_{1}}{{x}_{3}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}{{x}_{4}} \right)+{{x}_{3}}+{{x}_{4}}-\left( {{x}_{3}}{{x}_{1}}{{x}_{2}}+{{x}_{4}}{{x}_{1}}{{x}_{2}} \right)}{1-\left( {{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}} \right)+{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}-\left( {{x}_{1}}{{x}_{3}}+{{x}_{1}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{4}} \right)} \right) \\
& ={{\tan }^{-1}}\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}-\left( {{x}_{1}}{{x}_{3}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}{{x}_{4}}+{{x}_{3}}{{x}_{1}}{{x}_{2}}+{{x}_{4}}{{x}_{1}}{{x}_{2}} \right)}{1+{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}-\left( {{x}_{1}}{{x}_{2}}+{{x}_{3}}{{x}_{4}}+{{x}_{1}}{{x}_{3}}+{{x}_{1}}{{x}_{4}}+{{x}_{2}}{{x}_{3}}+{{x}_{2}}{{x}_{4}} \right)} \right) \\
\end{align}$
In the above expression substitute the values of sum of the roots, sum of the roots taken two at a time, sum of the roots taken three at a time and product of the roots from $eq.\left( 1,2,3,4 \right)$.
${{\tan }^{-1}}\left( \dfrac{\sin 2\beta -\cos \beta }{1-\sin \beta -\cos 2\beta } \right)$
In the above expression, we can write:
$\begin{align}
& \sin 2\beta =2\sin \beta \cos \beta \\
& 1-\cos 2\beta =2{{\sin }^{2}}\beta \\
\end{align}$
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{2\sin \beta \cos \beta -\cos \beta }{2{{\sin }^{2}}\beta -\sin \beta } \right) \\
& ={{\tan }^{-1}}\left( \dfrac{\cos \beta \left( 2\sin \beta -1 \right)}{\sin \beta \left( 2\sin \beta -1 \right)} \right) \\
\end{align}$
In the above expression $2\sin \beta -1$ will be cancelled out from the numerator and the denominator.
${{\tan }^{-1}}\left( \cot \beta \right)$
From the above solution, we have got the value of ${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$ as ${{\tan }^{-1}}\left( \cot \beta \right)$.
Note: Be careful in writing the formula of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-x\left( y \right)} \right)$. You might wrongly write the signs in this formula like in the numerator you can see a plus sign so instead of writing a plus sign, you could write minus sign and in the denominator, you can see a minus sign so instead of writing minus sign you can put plus sign.
Blunders in putting signs are very commonly observed.
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