# Proving analytic trig

• Mar 11th 2008, 03:36 PM
bilbobaggins
Proving analytic trig
1.
(1/(1+sinx)) + (1/(1-sinx)) = 2SEC(X)^2

2.1/(secx-tanx)= secx+tanx
• Mar 11th 2008, 03:42 PM
Moo
Hello,

For the first one, put the fractions of the left side on the same denominator. Then simplify with the cosē+sinē=1 formula ;)

For the second one, multiply the first term with 1-sin(x) on the numerator and denominator
• Mar 11th 2008, 10:20 PM
Soroban
Hello, bilbobaggins!

Quote:

$1)\;\;\frac{1}{1+\sin x} + \frac{1}{1-\sin x} \;=\;2\sec^2\!x$
$\frac{1}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} + \frac{1}{1-\sin x}\cdot{\color{blue}\frac{1+\sin x}{1+\sin x}} \;=\;\frac{1-\sin x}{1-\sin^2\!x} + \frac{1+\sin x}{1-\sin^2\!x}$

. . $= \;\frac{(1-\sin x) + (1 - \sin x)}{1-\sin^2\!x} \;=\;\frac{2}{\cos^2\!x} \;=\;2\sec^2\!x$

Quote:

$2)\;\;\frac{1}{\sec x-\tan x} \;=\;\sec x + \tan x$
$\frac{1}{\sec x - \tan x}\cdot{\color{blue}\frac{\sec x + \tan x}{\sec x + \tan x}} \;=\;\frac{\sec x + \tan x}{\underbrace{\sec^2\!x-\tan^2\!x}_{\text{This is 1}}} \;=\;\sec x + \tan x$