Answer
The identity is verified.
$\frac{sin~θ}{cos~θ}+\frac{cos~θ}{sin~θ}=csc~θ~sec~θ$
Work Step by Step
Use:
$\frac{1}{sin~θ}=csc~θ$
$\frac{1}{cos~θ}=sec~θ$
$sin^2θ+cos^2θ=1$
$\frac{sin~θ}{cos~θ}+\frac{cos~θ}{sin~θ}=\frac{sin~θ(sin~θ)}{cos~θ(sin~θ)}+\frac{cos~θ(cos~θ)}{sin~θ(cos~θ)}=\frac{sin^2θ}{sin~θ~cos~θ}+\frac{cos^2θ}{sin~θ~cos~θ}=\frac{sin^2θ+cos^2θ}{sin~θ~cos~θ}=\frac{1}{sin~θ~cos~θ}=\frac{1}{sin~θ}\frac{1}{cos~θ}=csc~θ~sec~θ$
