## Trigonometry (11th Edition) Clone

$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=2\sec^2\theta$$ The left side has been proved to be equal with the right side. It is thus an identity.
$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=2\sec^2\theta$$ The left side looks more complex, so we deal with the left side first. $$A=\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}$$ $$A=\frac{1+\sin\theta+1-\sin\theta}{(1-\sin\theta)(1+\sin\theta)}$$ $$A=\frac{2}{(1-\sin\theta)(1+\sin\theta)}$$ $$A=\frac{2}{1-\sin^2\theta}$$ (for $(a-b)(a+b)=a^2-b^2$) Now we can solve $1-\sin^2\theta$ by using a Pythagorean Identity: $$\cos^2\theta=1-\sin^2\theta$$ $$A=\frac{2}{\cos^2\theta}$$ From a Reciprocal Identity: $$\sec\theta=\frac{1}{\cos\theta}$$ So, $$\sec^2\theta=\frac{1}{\cos^2\theta}$$ Therefore, $$A=2\sec^2\theta$$ The trigonometric expression is hence an identity.