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}} $ is equal to

• A. $ \beta $
• B. $ \frac{\pi }{2}-\beta $
• C. $ \pi -\beta $
• D. $ -\beta $

Answer 1

Answer: (B)

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 $