1/sec + tan = sec - tan
prove one side equals to another BY SOLVING ONE SIDE ONLY
$\displaystyle \frac{1}{\sec \alpha+\tan \alpha}=\frac{1}{\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}}=\frac{\cos \alpha}{1+\sin \alpha} \cdot \frac{1-\sin \alpha}{1-\sin \alpha}=\frac{\cos \alpha \cdot(1-\sin \alpha)}{\cos^2 \alpha}=\frac{1-\sin \alpha}{\cos \alpha}=\frac{1}{\cos \alpha}-\tan \alpha=\sec \alpha-\tan \alpha$