Answer
$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$
The proof is in the Work Step by Step.
Work Step by Step
$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$
We take a look at the right side first.
$$X=\frac{1}{\sec t+\tan t}$$
$\sec t$ and $\tan t$ can be rewritten following the identities:
$$\sec t=\frac{1}{\cos t}\hspace{2cm}\tan t=\frac{\sin t}{\cos t}$$
So, $$X=\frac{1}{\frac{1}{\cos t}+\frac{\sin t}{\cos t}}$$
$$X=\frac{1}{\frac{1+\sin t}{\cos t}}$$
$$X=\frac{\cos t}{1+\sin t}$$
Now multiply both the numerator and denominator by $(1-\sin t)$:
$$X=\frac{\cos t(1-\sin t)}{(1+\sin t)(1-\sin t)}$$
$$X=\frac{\cos t(1-\sin t)}{1-\sin^2 t}$$ (do not forget that $(A-B)(A+B)=A^2-B^2$)
$$X=\frac{\cos t(1-\sin t)}{\cos^2t}$$(Pythagorean identity: $1-\sin^2 t=\cos^2t$)
$$X=\frac{1-\sin t}{\cos t}$$
Therefore the left and right sides are equal to each other.
$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$
The equation is an identity.
