Question

Cho góc \(\alpha\) thỏa mãn: \(\sin\alpha+\cos\alpha=\frac{\sqrt{2}}{2}\) . Tính P= \(\tan^2\alpha+\cot^2\alpha\)

Answer 1

\(sina+cosa=\frac{\sqrt{2}}{2}\Rightarrow sin^2a+cos^2a+2sina.cosa=\frac{1}{2}\)

\(\Rightarrow1+2sina.cosa=\frac{1}{2}\Rightarrow sina.cosa=\frac{-1}{4}\)

\(P=\frac{sin^2a}{cos^2a}+\frac{cos^2a}{sin^2a}=\frac{sin^4a+cos^4a}{\left(sina.cosa\right)^2}=\frac{\left(sin^2a+cos^2a\right)^2-2\left(sina.cosa\right)^2}{\left(sina.cosa\right)^2}\)

\(P=\frac{1-2.\left(\frac{-1}{4}\right)^2}{\left(-\frac{1}{4}\right)^2}=14\)