Answer
$$\frac{\csc t+1}{\csc t-1}=(\sec t+\tan t)^2$$
As the left side is equal to the right one, the equation is an identity.
Work Step by Step
$$\frac{\csc t+1}{\csc t-1}=(\sec t+\tan t)^2$$
1) We start from the left side.
$$X=\frac{\csc t+1}{\csc t-1}$$
- $\csc t$ can be rewritten into $\frac{1}{\sin t}$, according to reciprocal identities.
$$X=\frac{\frac{1}{\sin t}+1}{\frac{1}{\sin t}-1}$$
$$X=\frac{\frac{1+\sin t}{\sin t}}{\frac{1-\sin t}{\sin t}}$$
$$X=\frac{1+\sin t}{1-\sin t}$$
2) As the left side seems to not be able to simplify anymore, we expand the right side.
$$Y=(\sec t+\tan t)^2$$
- We rewrite $\sec t$ and $\tan t$ following the identities:
$$\sec t=\frac{1}{\cos t}\hspace{2cm}\tan t=\frac{\sin t}{\cos t}$$
$$Y=\Big(\frac{1}{\cos t}+\frac{\sin t}{\cos t}\Big)^2$$
$$Y=\Big(\frac{1+\sin t}{\cos t}\Big)^2$$
$$Y=\frac{(1+\sin t)^2}{\cos^2t}$$
- Pythagorean identity: $\cos^2t=1-\sin^2t=(1-\sin t)(1+\sin t)$
Therefore, $$Y=\frac{(1+\sin t)^2}{(1-\sin t)(1+\sin t)}$$
$$Y=\frac{1+\sin t}{1-\sin t}$$
So, $$X=Y=\frac{1+\sin t}{1-\sin t}$$
2 sides are thus equal, so the equation $$\frac{\csc t+1}{\csc t-1}=(\sec t+\tan t)^2$$ is an identity.
