## Trigonometry (11th Edition) Clone

$$1+\sin^2 x=\csc^2 x-\cot^2 x+\sin^2 x$$ $\text{D}$ is the answer.
$$A=1+\sin^2 x$$ Looking at this one, we can think of Pythagorean Identity $\sin^2 x+\cos^2 x=1$. However, it would change $A$ into $2\sin^2 x+\cos^2 x$, and there is no corresponding choice on the right side. So probably not this way. Until this exercise, if you solve each exercise continously, we will be left with choice B. $\frac{1}{\sec^2 x}$ and choice D. $\csc^2 x-\cot^2 x+\sin^2 x$ Choice D is actually quite complicated, and we can surely simplify it first. $$D=\csc^2 x-\cot^2 x+\sin^2 x$$ According to a Reciprocal Identity related to $\csc\theta$ and a Quotient Identity related to $\cot x$, $$\csc\theta=\frac{1}{\sin\theta}$$ $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ That means we can rewrite $D$ as follows: $$D=\frac{1}{\sin^2 x}-\frac{\cos^2 x}{\sin^2 x}+\sin^2 x$$ $$D=\frac{1-\cos^2 x}{\sin^2 x}+\sin^2 x$$ From Pythagorean Identities: $$\sin^2\theta+\cos^2\theta=1$$ $$\sin^2\theta=1-\cos^2\theta$$ Therefore, $$D=\frac{\sin^2 x}{\sin^2 x}+\sin^2 x$$ $$D=1+\sin^2 x$$ This coincidentally equals with $A$. So we can choose D right away. $\text{D}$ is the answer.