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

• May 29th 2008, 06:42 AM
pook
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
• May 29th 2008, 07:05 AM
angel.white
Quote:

Originally Posted by pook
help!!

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

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

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

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

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

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

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

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

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

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

$\displaystyle sec~x-tan~x= sec~x-tan~x$
• May 29th 2008, 07:07 AM
Jonboy
i can help with your first question.

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

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

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

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

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

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