Question

tính giá trị của biểu thức :

\(\tan^2\frac{\pi}{12}+\tan^2\frac{5\pi}{12}\)

Answer 1

\(=tan^2\frac{\pi}{12}+tan^2\left(\frac{\pi}{2}-\frac{\pi}{12}\right)=tan^2\frac{\pi}{12}+cot^2\frac{\pi}{12}\)

\(=tan^2\frac{\pi}{12}+cot^2\frac{\pi}{12}+2tan\frac{\pi}{12}.cot\frac{\pi}{12}-2\)

\(=\left(tan\frac{\pi}{12}+cot\frac{\pi}{12}\right)^2-2\)

\(=\left(\frac{1}{sin\frac{\pi}{12}.cos\frac{\pi}{12}}\right)^2-2\)

\(=\left(\frac{2}{2sin\frac{\pi}{12}.cos\frac{\pi}{12}}\right)^2-2\)

\(=\left(\frac{2}{sin\frac{\pi}{6}}\right)^2-2\)

\(=16-2=14\)