## Trigonometry (11th Edition) Clone

$$\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}=\cot\theta\csc\theta$$ We simplify the left side first, and the trigonometric expression is an identity.
$$\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}=\cot\theta\csc\theta$$ The left side is more complicated, which means we would deal with it first. $$A=\frac{\csc\theta+\cot\theta}{\tan\theta+\sin\theta}$$ We would use the following identities: $$\csc\theta=\frac{1}{\sin\theta}\hspace{1cm}\cot\theta=\frac{\cos\theta}{\sin\theta}\hspace{1cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Apply them into $A$: $$A=\frac{\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\sin\theta}$$ $$A=\frac{\frac{1+\cos\theta}{\sin\theta}}{\frac{\sin\theta+\sin\theta\cos\theta}{\cos\theta}}$$ $$A=\frac{1+\cos\theta}{\sin\theta}\times\frac{\cos\theta}{\sin\theta+\sin\theta\cos\theta}$$ $$A=\frac{1+\cos\theta}{\sin\theta}\times\frac{\cos\theta}{\sin\theta(1+\cos\theta)}$$ $$A=\frac{\cos\theta}{\sin^2\theta}$$ $$A=\frac{\cos\theta}{\sin\theta}\times\frac{1}{\sin\theta}$$ Also, from the following identities: $$\frac{\cos\theta}{\sin\theta}=\cot\theta\hspace{1.5cm}\frac{1}{\sin\theta}=\csc\theta$$ Therefore, $$A=\cot\theta\csc\theta$$ The trigonometric expression is therefore proved.