If {x_1}, {x_2}, {x_3}, {x_4} are roots of th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$. From Vieta's formulas,we have: $S_1 = \Sigma {x_1

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$. From Vieta's formulas,we have: $S_1 = \Sigma {x_1} = \sin 2\beta$ $S_2 = \Sigma {x_1}{x_2} = \cos 2\beta$ $S_3 = \Sigma {x_1}{x_2}{x_3} = \cos \beta$ $S_4 = {x_1}{x_2}{x_3}{x_4} = - \sin \beta$ We know that ${	an ^{ - 1}}{x_1} + {	an ^{ - 1}}{x_2} + {	an ^{ - 1}}{x_3} + {	an ^{ - 1}}{x_4} = {	an ^{ - 1}}\left( {\frac{{S_1 - S_3}}{{1 - S_2 + S_4}}} ight)$. Substituting the values: $= {	an ^{ - 1}}\left( {\frac{{\sin 2\beta - \cos \beta }}{{1 - \cos 2\beta - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{2\sin \beta \cos \beta - \cos \beta }}{{1 - (1 - 2\sin^2 \beta) - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{2\sin^2 \beta - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{\sin \beta (2\sin \beta - 1)}}} ight)$ $= {	an ^{ - 1}}(\cot \beta ) = {	an ^{ - 1}}\left[ {	an \left( {\frac{\pi }{2} - \beta } ight)} ight] = \frac{\pi }{2} - \beta $.

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} = $

Similar Questions

If $\sin^{-1} x = \frac{\pi}{5}$ for some $x \in (-1, 1)$,then the value of $\cos^{-1} x$ is

The value of $\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ is

If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$, then $5+x=$

$\cot \left[\sum_{n=3}^{32} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right]=$

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$,where $-1 < x < 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module