# How do you prove  (secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)?

Mar 15, 2018

To prove,we take
${\sec}^{2} A - 1 = {\tan}^{2} A , \mathmr{and} \tan A = \sin \frac{A}{\cos} A , \frac{1}{\cos} A = \sec A$

#### Explanation:

$\frac{\sec A - 1}{\sin} ^ 2 A = {\sec}^{2} \frac{A}{1 + \sec A}$
$L H S = \frac{\sec A - 1}{\sin} ^ 2 A = \frac{\left(\sec A - 1\right) \left(\sec A + 1\right)}{\left({\sin}^{2} A\right) \left(\sec A + 1\right)}$
$= \frac{{\sec}^{2} A - 1}{\left({\sin}^{2} A\right) \left(\sec A + 1\right)} = {\tan}^{2} \frac{A}{{\sin}^{2} A \left(\sec A + 1\right)} = \frac{{\sin}^{2} \frac{A}{\cos} ^ 2 A}{{\sin}^{2} A \left(\sec A + 1\right)} = \frac{\frac{1}{\cos} ^ 2 A}{\sec A + 1} = {\sec}^{2} \frac{A}{1 + \sec A} = R H S$

Mar 17, 2018

Kindly refer to a Proof in the Explanation.

#### Explanation:

We have, ${\sec}^{2} A - 1 = {\tan}^{2} A = {\sin}^{2} A \cdot \frac{1}{\cos} ^ 2 A$.

$\therefore \left(\sec A + 1\right) \left(\sec A - 1\right) = {\sin}^{2} A \cdot {\sec}^{2} A$.

 rArr (secA-1)/sin^2A=sec^2A/(1+secA, as desired!