Question

已知cosα=(m^2-n^2)/(m^2+n^2),求tanα和sinα的值

Answer 1

sin²α+cos²α=1

所以sin²α=1-cos²α=4m²n²/(m²+n²)²

则sinα=±2mn/(m²+n²)

则tanα=sinα/cosα=±2mn/(m²-n²)