Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 513: 50

$\frac{5}{tan~x+sec~x}=5(sec~x-tan~x)$

$\frac{5}{tan~x+sec~x}=\frac{5}{tan~x+sec~x}\frac{tan~x-sec~x}{tan~x-sec~x}=\frac{5~(tan~x-sec~x)}{tan^2x-sec^2x}=\frac{5~(tan~x-sec~x)}{tan^2x-(tan^2x+1)}=\frac{5~(tan~x-sec~x)}{-1}=5(sec~x-tan~x)$