## Trigonometry (11th Edition) Clone

$$\frac{1}{\sec t-1}+\frac{1}{\sec t+1}=2\cot t\csc t$$ The verification process is in the work step by step.
$$\frac{1}{\sec t-1}+\frac{1}{\sec t+1}=2\cot t\csc t$$ 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.