## Trigonometry (11th Edition) Clone

$$\frac{\csc t+1}{\csc t-1}=(\sec t+\tan t)^2$$ As the left side is equal to the right one, the equation is an identity.
$$\frac{\csc t+1}{\csc t-1}=(\sec t+\tan t)^2$$ 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.