Chọn đáp án đúng:
A. \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = \frac{1}{2}\tan \left( {x + \frac{\pi }{3}} \right)\tan \left( {\frac{\pi }{6} - x} \right)\)
B. \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = 2\tan \left( {x + \frac{\pi }{3}} \right)\tan \left( {\frac{\pi }{6} - x} \right)\)
C. \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = \tan \left( {x + \frac{\pi }{3}} \right)\tan \left( {\frac{\pi }{6} - x} \right)\)
D. \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = 3\tan \left( {x + \frac{\pi }{3}} \right)\tan \left( {\frac{\pi }{6} - x} \right)\)
Sử dụng công thức: \(\tan \left( {\frac{\pi }{2} - \alpha } \right) = \cot \alpha ,\tan \alpha .\cot \alpha = 1,{\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
Ta có: \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = \sqrt {{{\sin }^4}x + 4\left( {1 - {{\sin }^2}x} \right)} + \sqrt {{{\cos }^4}x + 4\left( {1 - {{\cos }^2}x} \right)} \)
\( = \sqrt {{{\left( {{{\sin }^2}x} \right)}^2} - 4{{\sin }^2}x + 4} + \sqrt {{{\left( {{{\cos }^2}x} \right)}^2} - 4{{\cos }^2}x + 4} = \sqrt {{{\left( {{{\sin }^2}x - 2} \right)}^2}} + \sqrt {{{\left( {{{\cos }^2}x - 2} \right)}^2}} \)
\( = \left( {2 - {{\sin }^2}x} \right) + \left( {2 - {{\cos }^2}x} \right) = 4 - \left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 3\)
Lại có: \(\left( {x + \frac{\pi }{3}} \right) + \left( {\frac{\pi }{6} - x} \right) = \frac{\pi }{2} \Rightarrow \tan \left( {\frac{\pi }{6} - x} \right) = \cot \left( {x + \frac{\pi }{3}} \right) \Rightarrow \tan \left( {\frac{\pi }{6} - x} \right).\tan \left( {x + \frac{\pi }{3}} \right) = 1\)
Do đó: \(\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = 3\tan \left( {x + \frac{\pi }{3}} \right)\tan \left( {\frac{\pi }{6} - x} \right)\)
Đáp án D
