\(\text{Ta có}:1+\text{tan}^2a=\frac{\text{kề}^2}{\text{kề}^2}+\frac{\text{đối}^2}{\text{kề}^2}=\frac{\text{kề}^2+\text{đối}^2}{\text{kề}^2}=\frac{\text{huyền}^2}{\text{kề}^2}=\frac{1}{\frac{\text{kề}^2}{\text{huyền}^2}}=\frac{1}{\text{cos}^2}\left(dpcm\right)\)