Trigonometry Identity Problem Prove that:
$$\frac{\tan A  + \sec A  - 1}{\tan A  - \sec A + 1}  =  \frac{1  +  \sin A}{ \cos A}$$
I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/cos A and then simplifying but it didn't work.
 A: Setting  $\displaystyle1=\sec^2A-\tan^2A$ in the numerator, 
So, the numerator becomes 
$$\tan A+\sec A-1$$
$$=\tan A+\sec A-(\sec^2A-\tan^2A)$$
$$=(\sec A+\tan A)\{1-(\sec A-\tan A)\}$$
$$=\frac{(1+\sin A)}{\cos A}(1-\sec A+\tan A)$$
A: You may also simplify the expression by using the definition of the tangent and the secant function:
\begin{align}
&\dfrac{\tan A +\sec A-1}{\tan A - \sec A +1}
\\=&\dfrac{\dfrac{\sin A}{\cos A} +\dfrac1{\cos A}-1}{\dfrac{\sin A}{\cos A} - \dfrac1{\cos A} +1}
\\=&\dfrac{\sin A -\cos A+1}{\sin A+\cos A-1}
\\=&\dfrac{(\sin A -\cos A+1)\cdot(\sin A+\cos A+1)}{(\sin A+\cos A-1)\cdot(\sin A+\cos A+1)}=\dfrac{(\sin A +1)^2-\cos^2 A}{(\sin A+\cos A)^2-1}
\\=&\dfrac{\sin^2 A+(1-\cos^2 A) +2\sin A}{2\sin A\cos A}\end{align}
The rest is fairly straightforward.
A: Multiplying the numerator & the denominator by $\cos A$
$$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$
$$\cos A(\sin A+1-\cos A)=\cos A\sin A+\cos A-\cos^2A$$
$$=\cos A(1+\sin A)-(1-\sin^2A)$$
$$\implies\cos A(\sin A+1-\cos A)=(1+\sin A)(\cos A-1+\sin A)$$
$$\implies\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=?$$
We can start with $(1+\sin A)(\sin A-1+\cos A)$ as well
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