Trig Identity help Watch

TheJ0ker
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#1
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If x = \sec \theta + \tan \theta show that \frac{1}{x} = \sec \theta - \tan \theta

I started by doing 1 over both sides of the first equation and ended up tidying the R.H.S to \frac{1}{x} = \frac{\cos \theta}{1 + \sin \theta} which I now can't do anything with. I'm assuming I have gone about the question wrong from the start but I can't see how I could start differently, help would be appreciated!

cheers.
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Mr M
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Do you know an identity connecting \sec^2 \theta and \tan^2 \theta ?
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aviva_123
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think about how you can rearrange the trigonometric identity: 1 + tan^2(\theta) = sec^2(\theta)
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TheJ0ker
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(Original post by Mr M)
Do you know an identity connecting \sec^2 \theta and \tan^2 \theta ?

(Original post by aviva_123)
think about how you can rearrange the trigonometric identity: 1 + tan^2(\theta) = sec^2(\theta)
I know now that I can basically rationalise the denominator of \frac{\cos \theta}{1+\sin \theta} and then solve it fairly easily, out of interest what method were you both suggesting? The only way I saw of being able to use 1 + \tan^2 \theta = \sec^2 \theta was to square everything and that didn't seem to help much....
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Mr M
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(Original post by TheJ0ker)
I know now that I can basically rationalise the denominator of \frac{\cos \theta}{1+\sin \theta} and then solve it fairly easily, out of interest what method were you both suggesting? The only way I saw of being able to use 1 + \tan^2 \theta = \sec^2 \theta was to square everything and that didn't seem to help much....
Difference of two squares?
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TheJ0ker
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(Original post by Mr M)
Difference of two squares?
Oh I see now, I would never have seen that by myself..... Cheeky trig identities.

Thanks for the help.
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