## Trigonometry (11th Edition) Clone

$$\sin^2\theta(1+\cot^2\theta)-1=0$$ The trigonometric expression is an identity.
$$\sin^2\theta(1+\cot^2\theta)-1=0$$ The left side would be simplified, since it is more complex. $$A=\sin^2\theta(1+\cot^2\theta)-1$$ From a Pythagorean Identity: $$1+\cot^2\theta=\csc^2\theta$$ Also, from a Reciprocal Identity: $$\csc\theta=\frac{1}{\sin\theta}$$ Therefore, $$1+\cot^2\theta=\Big(\frac{1}{\sin\theta}\Big)^2=\frac{1}{\sin^2\theta}$$ Apply back to $A$, we have: $$A=\sin^2\theta\times\frac{1}{\sin^2\theta}-1$$ $$A=1-1=0$$ The left side equals the right side, so it is an identity.