# Math Help - Trig Proof (+more)... VERY QUICKK PLEASE

1. ## Trig Proof (+more)... VERY QUICKK PLEASE

help!!

prove....
secX-tanX= 1/(secX+tanX)

then deduce that

0<secX - tanX <(or equal to) 1
for values of x between 0 and pi/2

fianlly..
given that secX- tanX = 1/2

find exact value of X

thanks

2. Originally Posted by pook
help!!

prove....
secX-tanX= 1/(secX+tanX)
$sec~x-tan~x= \frac 1{sec~x+tan~x}$

$sec~x-tan~x= \frac 1{\frac 1{cos~x}+\frac {sin~x}{cos~x}}$

$sec~x-tan~x= \frac 1{\frac {1+sin~x}{cos~x}}$

$sec~x-tan~x= \frac {cos~x}{1+sin~x}$

$sec~x-tan~x= \frac {cos~x}{1+sin~x}*\frac{1-sin~x}{1-sin~x}$

$sec~x-tan~x= \frac {cos~x-cos~x~sin~x}{1-sin^2~x}$

$sec~x-tan~x= \frac {cos~x-cos~x~sin~x}{cos^2~x}$

$sec~x-tan~x= \frac {cos~x}{cos^2~x}-\frac{cos~x~sin~x}{cos^2~x}$

$sec~x-tan~x= \frac {1}{cos~x}-\frac{sin~x}{cos~x}$

$sec~x-tan~x= sec~x-tan~x$

3. i can help with your first question.

$sec\,x - tan\,x = \frac{1}{sec\,x + tan\,x}$

Multiply both sides by $sec\,x + tan\,x$:

$sec^2\,x - tan^2\,x = 1$

There's an identity: $1 + tan^2\,x = sec^2\,x$

Re-arrange that to: $sec^2\,x - tan^2\,x = 1$

Thus the left side side of $sec^2\,x - tan^2\,x = 1$ becomes one and the proof is complete.