Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
 A: Notice, $$LHS=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}$$
$$=\frac{\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta}+1}$$
$$=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$
$$=\frac{(\sin\theta-\cos\theta+1)((\sin\theta+\cos\theta)+1)}{(\sin\theta+\cos\theta-1)((\sin\theta+\cos\theta)+1)}$$
$$=\frac{\sin^2\theta+2\sin \theta+1-\cos^2\theta}{(\sin\theta+\cos\theta)^2-1}$$
$$=\frac{\sin^2\theta+2\sin \theta+\sin^2\theta}{\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1}$$
$$=\frac{2\sin^2\theta+2\sin \theta}{1+2\sin\theta\cos\theta-1}$$
$$=\frac{2\sin\theta(1+\sin \theta)}{2\sin\theta\cos\theta}$$
$$=\frac{1+\sin \theta}{\cos\theta}=RHS$$
A: Let's start from the complicated side.
$$\begin{align*}
\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}&=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\times\frac{\cos\theta}{\cos\theta}\quad\text{(Multiply by $1=\frac{\cos\theta}{\cos\theta}$ to simplify)}\\
&=\frac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}+1\right)}\quad\text{(Forcefully factoring out the terms we need)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}\times\frac{1+\sin\theta}{1+\sin\theta}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin^2\theta-1}{\cos\theta(1+\sin\theta)}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{-\cos^2\theta}{\cos\theta(1+\sin\theta)}+1\right)}\\
&=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(1-\frac{\cos\theta}{1+\sin\theta}\right)}\\
&=\frac{1+\sin\theta}{\cos\theta}\quad\text{(Cancelling out common terms)}
\end{align*}$$
A: $$LHS = \frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta}$$
$$\Longleftrightarrow\frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta} = \frac{1+\sin\theta}{\cos\theta}$$
$$\Longleftrightarrow(\sin\theta+1-\cos\theta)\cos\theta = (\sin\theta - 1+ \cos\theta)(1+\sin\theta)$$
$$\Longleftrightarrow\sin\theta\cos\theta+\cos\theta-\cos^2\theta = \sin\theta - 1 + \cos\theta + \sin^2\theta-\sin\theta+\cos\theta\sin\theta$$
$$\Longleftrightarrow-\cos^2\theta = -1 + \sin^2\theta$$
Done.
A: 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}
