Question

(sec(x)+1)/tan(x)^2= 1/(sec(x)-1) identify identity

Answer 1

$$\frac {\sec (x) + 1}{\tan^ {2} (x)} = \frac {\frac {1}{\cos (x)} + 1}{\tan^ {2} (x) + 1 - 1} = \frac {\frac {1}{\cos (x)} + 1}{\frac {1}{\cos^ {2} (x)} - 1} = \frac {\frac {1}{\cos (x)} + 1}{(\frac {1}{\cos (x)} - 1) (\frac {1}{\cos (x)} + 1)} = \frac {1}{\frac {1}{\cos (x)} - 1} = \frac {1}{\sec (x) - 1}$$