# Question in a trigonometric equation. (1-sin A)/(1+ sin A) = 1 + 2 tan A (tan A - sec A )

Question in a trigonometric equation.
I tried it a lot but am not able to get this.Pls help in how should I think when solving this type of question and which side is better to try to simplify first (LHS or RHS).Please share the solution in that way.
$\frac{1-\mathrm{sin}A}{1+\mathrm{sin}A}=1+2\mathrm{tan}A\left(\mathrm{tan}A-\mathrm{sec}A\right)$
One of the ways I tried but am not understand that how to simplify it in such a way that you get the RHS?
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Brooklyn Mcintyre
Starting from the LHS we have
$\frac{1-\mathrm{sin}\left(A\right)}{1+\mathrm{sin}\left(A\right)}$
$=\frac{1-\mathrm{sin}\left(A\right)}{1+\mathrm{sin}\left(A\right)}\cdot \frac{1-\mathrm{sin}\left(A\right)}{1-\mathrm{sin}\left(A\right)}=\frac{1-2\mathrm{sin}\left(A\right)+{\mathrm{sin}}^{2}\left(A\right)}{1-{\mathrm{sin}}^{2}\left(A\right)}$
$=\frac{{\mathrm{cos}}^{2}\left(A\right)-2\mathrm{sin}\left(A\right)+2{\mathrm{sin}}^{2}\left(A\right)}{{\mathrm{cos}}^{2}\left(A\right)}$
using the identity ${\mathrm{sin}}^{2}\left(A\right)+{\mathrm{cos}}^{2}\left(A\right)=1$. Can you end it now?