Question

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 ${\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4=$

• A. $\beta$
• B. $\left(\pi /2\right)-\beta$
• C. $\mathrm{\pi }-\mathrm{\beta }$
• D. $-\beta$

Answer 1

The correct option is B

$\left(\pi /2\right)-\beta$

Explanation for the correct option :

From the given equation, $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =0$, we have

$\begin{array}{rcl}\Sigma x_1& =& \sin 2\beta \\ \Sigma x_1{x}_2& =& \cos 2\beta \\ \Sigma x_1{x}_2{x}_3& =& \cos \beta \\ x_1{x}_2{x}_3{x}_4& =& -\sin \beta \end{array}$

Now,

$\begin{array}{rcl}& & {\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4\\ & =& {\tan }^{-1}\left[\frac{\Sigma x_1-\Sigma x_1{x}_2{x}_3}{1-\Sigma x_1{x}_2+x_1{x}_2{x}_3{x}_4}\right]\\ & =& {\tan }^{-1}\left[\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta }\right]\\ & =& {\tan }^{-1}\left[\frac{2\sin \beta \cos \beta -\cos \beta }{1-\left(1-2{\sin }^2\beta \right)-\sin \beta }\right]\left[by\sin 2x=2s\mathrm{in}x\cos x,and\cos 2x=1-2{\sin }^{2}x\right]\\ & =& {\tan }^{-1}\left[\frac{\cos \beta \left(2\sin \beta -1\right)}{2{\sin }^2\beta -\sin \beta }\right]\\ & =& {\tan }^{-1}\left[\frac{\cos \beta \left(2\sin \beta -1\right)}{\sin \beta \left(2\sin \beta -1\right)}\right]\\ & =& {\tan }^{-1}\left[\cot \beta \right]\\ & =& {\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{2}-\beta \right)\right]\\ & =& \frac{\mathrm{\pi }}{2}-\beta \end{array}$

Hence, option B is correct.