## Trigonometry (11th Edition) Clone

$$\csc^2\theta+\sec^2\theta=\frac{1}{\sin^2\theta\cos^2\theta}$$
$$A=\csc^2\theta+\sec^2\theta$$ $\csc\theta$ and $\sec\theta$ can be written according to the following identities: $$\csc\theta=\frac{1}{\sin\theta}\hspace{2cm}\sec\theta=\frac{1}{\cos\theta}$$ So, $$A=\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}$$ $$A=\frac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}$$ - Recall the identity $\cos^2\theta+\sin^2\theta=1$ $$A=\frac{1}{\sin^2\theta\cos^2\theta}$$ This is the answer we need to the question.