Question

If $x_1,x_2,x_3,$ and $x_4$ 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. $\frac{\mathrm{\pi }}{2}-\beta$
• C. $\mathrm{\pi }-\mathrm{\beta }$
• D. $-\beta$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{2}-\beta$

Explanation for the correct option:

Step -1: Find the sum and product of roots:

For the equation $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =0$, the roots of the equation are $x_1,x_2,x_3,$ and $x_4$ .

Now, the sum of roots is given as: $S_1=x_1+x_2+x_3+x_4=-(-\sin 2\beta )=\sin 2\beta$.

The sum of product of two distinct roots is given as: $S_2=\sum _{}{x}_i{x}_j=\cos 2\beta$.

The sum of product of three distinct roots is given as: $S_3=\sum _{}{x}_i{x}_j{x}_k=-(-\cos \beta )=\cos \beta$.

The product of roots is given as: $S_4=x_1{x}_2{x}_3{x}_4=-\sin \beta$.

Step- 2: Find the value of the expression:

The sum of inverse tangent functions ${\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4$:

${\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4={\tan }^{-1}\left(\frac{S_1-S_3}{1-S_2+S_4}\right)$.

Substitute $S_1=\sin 2\beta ,S_2=\cos 2\beta ,S_3=\cos \beta ,S_4=-\sin \beta$ and find its value.

$\begin{array}{rcl}{\tan }^{-1}\left(\frac{S_1-S_3}{1-S_2+S_4}\right)& =& {\tan }^{-1}\left(\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta +(-\sin \beta )}\right)\\ & =& {\tan }^{-1}\left(\frac{2\sin \beta \cos \beta -\cos \beta }{2{\sin }^2\beta -\sin \beta }\right)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{A}\mathbf{}\mathbf{;}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{A}\mathbf{]}\\ & =& {\tan }^{-1}\left(\frac{\cos \beta (2\sin \beta -1)}{\sin \beta (2\sin \beta -1)}\right)\\ & =& {\tan }^{-1}\left(\cot \beta \right)\\ & =& {\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{2}-\beta \right)\right)\\ & =& \frac{\mathrm{\pi }}{2}-\beta \mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\left(tanx\right)\mathbf{=}\mathbf{x}\mathbf{]}\end{array}$

Thus the value of ${\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4$ is $\frac{\mathrm{\pi }}{2}-\beta$.

Hence, the correct option is B.