Question

tan/sec+1+sec+1/tan = 2cosec​

Answer 1

$\frac{tan}{sec + 1} + \frac{sec + 1}{tan} \\ \\ = \frac{ {tan}^{2} + ( {sec + 1})^{2} }{tan(sec + 1)} \\ \\ = \frac{ {tan}^{2} + {sec}^{2} + 1 + 2sec}{tan(sec + 1)} \\ \\ = \frac{2 {sec}^{2} + 2sec }{tan(sec +1 )} \\ \\ = \frac{2sec(sec + 1}{tan(sec + 1)} \\ \\ = 2 \frac{sec}{tan} \\ \\ = 2cosec$