Question

$\frac{\mathrm{cosec}A-\sin A}{\mathrm{cosec}A+\sin A}=\frac{{\sec }^{2}A-{\tan }^{2}A}{{\sec }^{2}A+{\tan }^{2}A}$

Answer 1

$$\begin{gathered} \frac{\mathrm{cosec}A-\sin A}{\mathrm{cosec}A+\sin A} \\ =\frac{{\displaystyle \frac{1}{\sin A}}-\sin A}{{\displaystyle \frac{1}{\sin A}}+\sin A} \\ =\frac{{\displaystyle \frac{1-{\sin }^{2}A}{\sin A}}}{{\displaystyle \frac{1+{\sin }^{2}A}{\sin A}}} \\ =\frac{1-{\sin }^{2}A}{1+{\sin }^{2}A} \end{gathered}$$
$=\frac{{\displaystyle \frac{1-{\sin }^{2}A}{{\cos }^{2}A}}}{{\displaystyle \frac{1+{\sin }^{2}A}{{\cos }^{2}A}}}$ (Dividing numerator and denominator by cos²A)
$$\begin{gathered} =\frac{{\displaystyle \frac{1}{{\cos }^{2}A}-\frac{{\sin }^{2}A}{{\cos }^{2}A}}}{{\displaystyle \frac{1}{{\cos }^{2}A}+\frac{{\sin }^{2}A}{{\cos }^{2}A}}} \\ =\frac{{\sec }^{2}A-{\tan }^{2}A}{{\sec }^{2}A+{\tan }^{2}A}\left(\sec \theta =\frac{1}{\cos \theta }\mathrm{and}\tan \theta =\frac{\sin \theta }{\cos \theta }\right) \end{gathered}$$