Chapter 7 - 7.2 - Verifying Trigonometric Identities - 7.2 Exercises - Page 521: 44

$sec(-θ)\ne-sec~θ$ $sin(-θ)\ne sin~θ$ $\frac{1+sec(-θ)}{sin(-θ)+tan(-θ)}=-csc~θ$

$sec(-θ)=sec~θ$ $sin(-θ)=-sin~θ$ $tan(-θ)=-tan~θ$ $\frac{1+sec(-θ)}{sin(-θ)+tan(-θ)}=\frac{1+sec~θ}{-sin~θ-tan~θ}=\frac{1+sec~θ}{-sin~θ~(1+\frac{1}{cos~θ})}=\frac{1+sec~θ}{-sin~θ~(1+sec~θ)}=\frac{1}{-sin~θ}=-csc~θ$