<s>[INST] Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
 [/INST] Multiplying by $\cos\theta$ the num and the denom
$$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1} = \frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}$$
Now setting the equality and deleting the fractions
\begin{align}
\frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}&=\frac{1+\sin \theta}{\cos \theta}\\
\cos \theta + \cos \theta - \cos^2  \theta &= \sin  \theta + \sin^2  \theta-\sin  \theta-1+ \cos  \theta+\sin \theta \cos  \theta\\
-\cos^2  \theta &=\sin^2  \theta-1.
\end{align}
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