Question

$\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}={\left(\sec \mathrm{\theta }-\tan \mathrm{\theta }\right)}^2$

Answer 1

$$\begin{gathered} \frac{1-\sin \theta }{1+\sin \theta } \\ =\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta } \\ =\frac{{\left(1-\sin \theta \right)}^2}{\left(1+\sin \theta \right)\left(1-\sin \theta \right)} \\ =\frac{{\left(1-\sin \theta \right)}^2}{1-{\sin }^2\theta }\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \end{gathered}$$
$$\begin{gathered} =\frac{{\left(1-\sin \theta \right)}^2}{{\cos }^2\theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ ={\left(\frac{1-\sin \theta }{\cos \theta }\right)}^2 \\ ={\left(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }\right)}^2 \\ ={\left(\sec \theta -\tan \theta \right)}^2 \end{gathered}$$