Answer
$$\frac{1}{\sec t-1}+\frac{1}{\sec t+1}=2\cot t\csc t$$
The verification process is in the work step by step.
Work Step by Step
$$\frac{1}{\sec t-1}+\frac{1}{\sec t+1}=2\cot t\csc t$$
From the left side:
$$X=\frac{1}{\sec t-1}+\frac{1}{\sec t+1}$$
$$X=\frac{\sec t+1+\sec t-1}{(\sec t-1)(\sec t+1)}$$
$$X=\frac{2\sec t}{\sec^2t-1}$$
- Reciprocal identity: $\sec t=\frac{1}{\cos t}$
$$X=\frac{\frac{2}{\cos t}}{\frac{1}{\cos^2t}-1}$$
$$X=\frac{\frac{2}{\cos t}}{\frac{1-\cos^2t}{\cos^2t}}$$
$$X=\frac{2\cos^2t}{\cos t(1-\cos^2t)}$$
$$X=\frac{2\cos t}{1-\cos^2 t}$$
- Pythagorean identity: $1-\cos^2t=\sin^2t$
$$X=\frac{2\cos t}{\sin^2t}$$
Next, from the right side:
$$Y=2\cot t\csc t$$
- Using the identities: $$\cot t=\frac{\cos t}{\sin t}\hspace{2cm}\csc t=\frac{1}{\sin t}$$
$$Y=2\times\frac{\cos t}{\sin t}\times\frac{1}{\sin t}$$
$$Y=\frac{2\cos t}{\sin^2t}$$
Therefore $X=Y$, so $$\frac{1}{\sec t-1}+\frac{1}{\sec t+1}=2\cot t\csc t$$
The equation is an identity.
