# Thread: Prove the identity plz

1. ## Prove the identity plz

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

2. Originally Posted by nightrider456
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
first of all

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$

the way you have this posted is ambiguous (but looks like an easy solution somewhere)

3. ## solution

$\displaystyle \frac 1 {sec(x) + tan(x)} = \frac 1 { \frac 1 {cos(x)} + \frac {sin(x)} {cos(x)} } = \frac {cos(x)} {1 + sin(x)} = \frac {cos(x)} {1 + sin(x)} \cdot \frac {1 - sin(x)} {1 - sin(x)} =$

$\displaystyle \frac {cos(x) - cos(x)sin(x) } {cos^2(x)} = \frac {1 - sin(x)} {cos(x)}$

4. $\displaystyle \frac{1}{\sec{x}+\tan{x}} \cdot \frac{\sec{x}-\tan{x}}{\sec{x}-\tan{x}} =$

$\displaystyle \frac{\sec{x} - \tan{x}}{\sec^2{x} - \tan^2{x}} =$

$\displaystyle \sec{x} - \tan{x} =$

$\displaystyle \frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}} =$

$\displaystyle \frac{1-\sin{x}}{\cos{x}}$