# Find the value of (secx-1)(secx+1) and 1-sin^2theta/tan^2theta?

Jul 4, 2016

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

#### Explanation:

$\left(\sec x - 1\right) \left(\sec x + 1\right) = {\sec}^{2} x - 1 = 1 + {\tan}^{2} x - 1 = {\tan}^{2} x$

$1 - {\sin}^{2} \frac{\theta}{\tan} ^ 2 \theta = 1 - {\sin}^{2} \frac{\theta}{{\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta}$

= $1 - {\sin}^{2} \theta \times {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

= $1 - {\cos}^{2} \theta$

= ${\sin}^{2} \theta$