已知tanα=2,求1/（sin^2α－sinαcosα－cos^α）.
由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/（sin^2α－sinαcosα－cos^α）=1/（sin^2α－3cos^α）=1/(4/5－3/5)=5 想问的是：①sin^2α=4/5,cos^2α=1/5是怎么得出来的?②1/（sin^2α－sinαcosα－cos^α）是怎么得到1/（sin^2α－3cos^α）的?