## Trigonometry (11th Edition) Clone

$$\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}$$ The equation has been verified to be an identity.
$$\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}$$ We would deal with the right side first, taking $\sec\theta=\frac{1}{\cos\theta}$ $$\frac{\cos\theta}{\sec\theta-1}$$ $$=\frac{\cos\theta}{\frac{1}{\cos\theta}-1}$$ $$=\frac{\cos\theta}{\frac{1-\cos\theta}{\cos\theta}}$$ $$=\frac{\cos^2\theta}{1-\cos\theta}$$ Now we deal with the left side, but we already have something in mind. Since the right side can be transformed all to $\cos\theta$, we would try to do the same with the left side. For $\tan\theta$, we have $\tan\theta=\frac{\sin\theta}{\cos\theta}$ $$\frac{\cos\theta+1}{\tan^2\theta}$$ $$=\frac{\cos\theta+1}{\frac{\sin^2\theta}{\cos^2\theta}}$$ $$=\frac{\cos^2\theta(\cos\theta+1)}{\sin^2\theta}$$ With $\sin^2\theta$, we would apply $\sin^2\theta=1-\cos^2\theta$ $$=\frac{\cos^2\theta(\cos\theta+1)}{1-\cos^2\theta}$$ Now we simplify $$=\frac{\cos^2\theta(\cos\theta+1)}{(1-\cos\theta)(1+\cos\theta)}$$ $$=\frac{\cos^2\theta}{1-\cos\theta}$$ Both sides are then equal. $$\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}$$ The equation has been verified to be an identity.