## Trigonometry (11th Edition) Clone

$$\frac{1+\sin\theta}{1-\sin\theta}-\frac{1-\sin\theta}{1+\sin\theta}=4\tan\theta\sec\theta$$ 2 sides are equal, so the expression is an identity.
$$\frac{1+\sin\theta}{1-\sin\theta}-\frac{1-\sin\theta}{1+\sin\theta}=4\tan\theta\sec\theta$$ We simplify the left side first, since it is obviously more complex-looking. $$A=\frac{1+\sin\theta}{1-\sin\theta}-\frac{1-\sin\theta}{1+\sin\theta}$$ $$A=\frac{(1+\sin\theta)^2-(1-\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}$$ (we know that $a^2-b^2=(a-b)(a+b)$) $$A=\frac{[(1+\sin\theta)-(1-\sin\theta)][(1+\sin\theta)+(1-\sin\theta)]}{1-\sin^2\theta}$$ $$A=\frac{[1+\sin\theta-1+\sin\theta][1+\sin\theta+1-\sin\theta]}{1-\sin^2\theta}$$ $$A=\frac{2\sin\theta\times2}{1-\sin^2\theta}$$ $$A=\frac{4\sin\theta}{\cos^2\theta}$$ (from Pythagorean Identity: $1-\sin^2\theta=\cos^2\theta$) $$A=4\frac{\sin\theta}{\cos\theta}\frac{1}{\cos\theta}$$ We would apply the identities: $\frac{\sin\theta}{\cos\theta}=\tan\theta$ and $\frac{1}{\cos\theta}=\sec\theta$. That means, $$A=4\tan\theta\sec\theta$$ The expression is hence proved to be an identity, since both sides are equal.