Cho \(\tan\alpha+\cot\alpha=m\) với \(m>2\). Khi đó \(\left|\tan\alpha-\cot\alpha\right|\) bằng
\(\sqrt{4-m^2}\).\(\sqrt{m-4}\).\(\sqrt{4-m}\).\(\sqrt{m^2-4}\).Hướng dẫn giải:\(\tan\alpha+\cot\alpha=m\) \(\Rightarrow\left(\tan\alpha+\cot\alpha\right)^2=m^2\) \(\Rightarrow\tan^2\alpha+\cot^2\alpha+2.\tan\alpha.\cot\alpha=m^2\)
\(\Rightarrow\tan^2\alpha+\cot^2\alpha+2=m^2\Rightarrow\) \(\tan^2\alpha+\cot^2\alpha=m^2-2\)
Có \(\left(\tan\alpha-\cot\alpha\right)^2=\tan^2\alpha+\cot^2\alpha-2.\tan\alpha.\cot\alpha\) \(=m^2-2-2=m^2-4\)
Nên \(\left|\tan\alpha-\cot\alpha\right|\) = \(\sqrt{m^2-4}\).
