# Solve the equation frac{1-sin x}{1+sin x}=(sec x-tan x)^{2}

Question
Solve the equation $$\frac{1-\sin x}{1+\sin x}=(\sec x-\tan x)^{2}$$

2020-10-29
$$(\sec(x)-\tan(x))^{2}=\bigg(\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}\bigg)^{2}$$
$$=\bigg(\frac{1-\sin(x)}{\cos(x)}\bigg)^{2}$$
$$=\frac{(1-\sin(x))^{2}}{cos^{2}(x)}$$
$$=\frac{(1-\sin(x))^{2}}{1-sin^2(x)}$$
$$= \frac{-(1-\sin(x))^{2}}{(\sin(x)+1)(\sin(x)-1)}$$
$$=\frac{1-\sin(x)}{\sin(x)+1}$$
$$=\frac{1-\sin(x)}{1+\sin(x)}$$
Therefore
$$\frac{1-\sin(x)}{1+\sin(x)}=(\sec(x)-\tan(x))^{2}$$

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