1. ## proof

Given that $\displaystyle x = sec\theta + tan\theta$

show that $\displaystyle \frac{1}{x} = sec\theta - tan\theta$

2. $\displaystyle \frac{1}{x}=\frac{1}{\displaystyle\frac{1}{\cos\th eta}+\frac{\sin\theta}{\cos\theta}}=\frac{\cos\the ta}{1+\sin\theta}=$

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

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

3. Just been thinking, is this method correct?

$\displaystyle x = sec\theta + tan\theta$

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

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

multiply both sides by $\displaystyle sec\theta+tan\theta$

$\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta )$

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

$\displaystyle 1 = x(sec\theta - tan\theta)$

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

4. This exact question was posted yesterday here : http://www.mathhelpforum.com/math-he...-identity.html

5. Originally Posted by Tweety
Just been thinking, is this method correct?

$\displaystyle x = sec\theta + tan\theta$

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

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

multiply both sides by $\displaystyle sec\theta+tan\theta$

$\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta )$

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

$\displaystyle 1 = x(sec\theta - tan\theta)$

$\displaystyle \frac{1}{x} = sec\theta - tan\theta$
You seem to be assuming that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$ and that $\displaystyle {x} = \sec\theta + \tan\theta$, doing a few operations on these premises, undoing those operations, and then concluding that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$.

6. Originally Posted by Bruno J.
This exact question was posted yesterday here : http://www.mathhelpforum.com/math-he...-identity.html

Well than they must be using the same excercise book as me, as my exam board is with edexcel, this is the book I am using!

Edexcel AS and A Level Modular Mathematics - Core Mathematics 3: Amazon.co.uk: Mr Keith Pledger: Books

7. Originally Posted by Bruno J.
You seem to be assuming that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$ and that $\displaystyle {x} = \sec\theta + \tan\theta$, doing a few operations on these premises, undoing those operations, and then concluding that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$.

Well yes becasue thats what the book says?

can you tell me what steps are not 'valid'?

Also when I have the expression $\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta )$ and than divide most sides by $\displaystyle sec\theta - tan\theta$

would that not cancel $\displaystyle sec\theta - tan\theta$ from the right hand side? so I get this expression; $\displaystyle \frac{1}{sec\theta - tan\theta} = sec\theta+tan\theta$

thanks.

8. No, you are given that $\displaystyle {x} = \sec\theta + \tan\theta$, from which you must deduce that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$.

You cannot use in your proof that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$ because that is what you are trying to show. However that is what you seem to be doing in the third line : starting from $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$, you replaced $\displaystyle x$ by $\displaystyle \sec\theta + \tan\theta$.

9. Originally Posted by Bruno J.
No, you are given that $\displaystyle {x} = \sec\theta + \tan\theta$, from which you must deduce that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$.

You cannot use in your proof that $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$ because that is what you are trying to show. However that is what you seem to be doing in the third line : starting from $\displaystyle \frac{1}{x} = \sec\theta - \tan\theta$, you replaced $\displaystyle x$ by $\displaystyle \sec\theta + \tan\theta$.

oh I get what you meant now,

Thanks