## Trigonometry (11th Edition) Clone

$$\frac{1-\sin\theta}{1+\sin\theta}=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$ The trigonometric expression is an identity.
$$\frac{1-\sin\theta}{1+\sin\theta}=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$ We would deal with the right side first. $$A=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$ We can use $$a^2-2ab+b^2=(a-b)^2$$ That means $$A=(\sec\theta-\tan\theta)^2$$ Now we transform $\sec\theta$ and $\tan\theta$ according to the identities: $$\sec\theta=\frac{1}{\cos\theta}\hspace{2cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Therefore $A$ would be $$A=\Big(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}\Big)^2$$ $$A=\Big(\frac{1-\sin\theta}{\cos\theta}\Big)^2$$ $$A=\frac{(1-\sin\theta)^2}{\cos^2\theta}$$ From Pythagorean Identity: $\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$ That makes $A$ into $$A=\frac{(1-\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}$$ $$A=\frac{1-\sin\theta}{1+\sin\theta}$$ The left side is therefore equal with the right side. The trigonometric expression is thus an identity.