A) \[\beta \]
B) \[\frac{\pi }{2}-\beta \]
C) \[\pi -\beta \]
D) \[-\beta \]
Correct Answer: B
Solution :
We have, \[\Sigma {{x}_{1}}=\sin 2\beta ,\Sigma {{x}_{1}}{{x}_{2}}=\cos 2\beta \] \[\Sigma {{x}_{1}}{{x}_{2}}{{x}_{3}}=\cos \beta \]and\[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=-\sin \beta \] \[\therefore \]\[{{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}\] \[={{\tan }^{-1}}\left( \frac{\Sigma {{x}_{1}}-\Sigma {{x}_{1}}{{x}_{2}}{{x}_{3}}}{1-\Sigma {{x}_{1}}{{x}_{2}}+{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}} \right)\] \[={{\tan }^{-1}}\left( \frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta } \right)\] \[={{\tan }^{-1}}\left( \frac{(2\sin \beta -1)\cos \beta }{\sin \beta (2\sin \beta -1)} \right)={{\tan }^{-1}}(\cot \beta )\] \[={{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{2}-\beta \right) \right]=\frac{\pi }{2}-\beta \]
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