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

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

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- Aug 10th 2009, 10:39 AMTweetyproof
Given that $\displaystyle x = sec\theta + tan\theta $

show that $\displaystyle \frac{1}{x} = sec\theta - tan\theta $ - Aug 10th 2009, 10:44 AMred_dog
$\displaystyle \frac{1}{x}=\frac{1}{\displaystyle\frac{1}{\cos\th eta}+\frac{\sin\theta}{\cos\theta}}=\frac{\cos\the ta}{1+\sin\theta}=$

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

$\displaystyle =\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}=\sec\theta-\tan\theta$ - Aug 10th 2009, 10:55 AMTweety
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 $ - Aug 10th 2009, 10:56 AMBruno J.
This exact question was posted yesterday here : http://www.mathhelpforum.com/math-he...-identity.html

- Aug 10th 2009, 11:00 AMBruno 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

$. - Aug 10th 2009, 11:02 AMTweety

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 - Aug 10th 2009, 11:23 AMTweety

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. - Aug 10th 2009, 02:29 PMBruno 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

$. - Aug 11th 2009, 02:20 AMTweety