## Trigonometry (11th Edition) Clone

$$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$ The proof is in the 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.