## Trigonometry (11th Edition) Clone

$$\frac{(\sec\theta-\tan\theta)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}=2\tan\theta$$ The equation is an identity.
$$\frac{(\sec\theta-\tan\theta)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}=2\tan\theta$$ Obviously, the left side is so complicated compared with the right one that we'd better simplify it first. $$\frac{(\sec\theta-\tan\theta)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}$$ $$=\frac{\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}$$ We notice $\tan^2\theta+1=\sec^2\theta$, which means $$=\frac{\sec^2\theta-2\sec\theta\tan\theta+\sec^2\theta}{\sec\theta\csc\theta-\tan\theta\csc\theta}$$ $$=\frac{2\sec^2\theta-2\sec\theta\tan\theta}{\csc\theta(\sec\theta-\tan\theta)}$$ $$=\frac{2\sec\theta(\sec\theta-\tan\theta)}{\csc\theta(\sec\theta-\tan\theta)}$$ $$=\frac{2\sec\theta}{\csc\theta}$$ $\sec\theta=\frac{1}{\cos\theta}$ and $\csc\theta=\frac{1}{\sin\theta}$. Therefore, $$=2\frac{\frac{1}{\cos\theta}}{\frac{1}{\sin\theta}}$$ $$=2\frac{\sin\theta}{\cos\theta}$$ $$=2\tan\theta$$ The equation is therefore an identity.