Solve this

Question:

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

Solution:

$\frac{\operatorname{cosec} A-\sin A}{\operatorname{cosec} A+\sin A}$

$=\frac{\frac{1}{\sin A}-\sin A}{\frac{1}{\sin A}+\sin A}$

$=\frac{\frac{1-\sin ^{2} A}{\sin A}}{\frac{1+\sin ^{2} A}{\sin A}}$

$=\frac{1-\sin ^{2} A}{1+\sin ^{2} A}$

$=\frac{\frac{1-\sin ^{2} A}{\cos ^{2} A}}{\frac{1+\sin ^{2} A}{\cos ^{2} A}}$   (Dividing numerator and denominator by cos2A)

$=\frac{\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A}{\cos ^{2} A}}{\frac{1}{\cos ^{2} A}+\frac{\sin ^{2} A}{\cos ^{2} A}}$

$=\frac{\sec ^{2} A-\tan ^{2} A}{\sec ^{2} A+\tan ^{2} A} \quad\left(\sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\tan \theta=\frac{\sin \theta}{\cos \theta}\right)$

 

 

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