Trigonometry (11th Edition) Clone

$$\frac{1+\sin\theta}{\cot^2\theta}=\frac{\sin\theta}{\csc\theta-1}$$ The equation is an identity.
$$\frac{1+\sin\theta}{\cot^2\theta}=\frac{\sin\theta}{\csc\theta-1}$$ We examine from the left side first. $$A=\frac{1+\sin\theta}{\cot^2\theta}$$ $\cot\theta$ can be written according to Quotient Identity, in which $\cot\theta=\frac{\cos\theta}{\sin\theta}$ $$A=\frac{1+\sin\theta}{\frac{\cos^2\theta}{\sin^2\theta}}$$ $$A=\frac{\sin^2\theta(1+\sin\theta)}{\cos^2\theta}$$ Next, we write $\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$, following the Pythagorean Identity. $$A=\frac{\sin^2\theta(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$$ $$A=\frac{\sin^2\theta}{1-\sin\theta}$$ The left side can be temporarily stopped being examined to switch to the right side. $$B=\frac{\sin\theta}{\csc\theta-1}$$ $\csc\theta$ can be written according to Reciprocal Identity: $\csc\theta=\frac{1}{\sin\theta}$ $$B=\frac{\sin\theta}{\frac{1}{\sin\theta}-1}$$ $$B=\frac{\sin\theta}{\frac{1-\sin\theta}{\sin\theta}}$$ $$B=\frac{\sin^2\theta}{1-\sin\theta}$$ Thus, $A=B$. 2 sides are equal, so the equation is verified to be an identity.