Maths Question:
$\text{Verify }\sec 2m=\frac{{{\sec }^{2}}m}{2-{{\sec }^{2}}m}$
Maths Solution:
$\begin{align} & \sec 2m=\frac{1}{\cos 2m}=\frac{1}{{{\cos }^{2}}m-{{\sin }^{2}}m} \\ & \text{Divide both numerator and denominator with }{{\cos }^{2}}m \\ & \sec 2m=\frac{\tfrac{1}{{{\cos }^{2}}m}}{\tfrac{{{\cos }^{2}}m}{{{\cos }^{2}}m}-\tfrac{{{\sin }^{2}}m}{{{\cos }^{2}}m}}=\frac{{{\sec }^{2}}m}{1-{{\tan }^{2}}m} \\ & \sec 2m=\frac{{{\sec }^{2}}m}{1-({{\sec }^{2}}m-1)}=\frac{{{\sec }^{2}}m}{2-{{\sec }^{2}}m} \\\end{align}$
University mathstopic:
