Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$

$$LHS = \frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta}$$
$$\Longleftrightarrow\frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta} = \frac{1+\sin\theta}{\cos\theta}$$
$$\Longleftrightarrow(\sin\theta+1-\cos\theta)\cos\theta = (\sin\theta - 1+ \cos\theta)(1+\sin\theta)$$
$$\Longleftrightarrow\sin\theta\cos\theta+\cos\theta-\cos^2\theta = \sin\theta - 1 + \cos\theta + \sin^2\theta-\sin\theta+\cos\theta\sin\theta$$
$$\Longleftrightarrow-\cos^2\theta = -1 + \sin^2\theta$$
Done.
