I have been working on this problem for quite some time and have gotten to the point that I am stuck. any way to prove it would be greatly appreciated.
1
secx+tanx
=
1-sinx
cosx
I have been working on this problem for quite some time and have gotten to the point that I am stuck. any way to prove it would be greatly appreciated.
1
secx+tanx
=
1-sinx
cosx
$\displaystyle
\frac 1 {sec(x) + tan(x)} = \frac 1 { \frac 1 {cos(x)} + \frac {sin(x)} {cos(x)} } = \frac {cos(x)} {1 + sin(x)} = \frac {cos(x)} {1 + sin(x)} \cdot \frac {1 - sin(x)} {1 - sin(x)} =
$
$\displaystyle
\frac {cos(x) - cos(x)sin(x) } {cos^2(x)} = \frac {1 - sin(x)} {cos(x)}
$
$\displaystyle \frac{1}{\sec{x}+\tan{x}} \cdot \frac{\sec{x}-\tan{x}}{\sec{x}-\tan{x}} = $
$\displaystyle \frac{\sec{x} - \tan{x}}{\sec^2{x} - \tan^2{x}} =$
$\displaystyle \sec{x} - \tan{x} = $
$\displaystyle \frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}} =
$
$\displaystyle \frac{1-\sin{x}}{\cos{x}}$