Once you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)

Expand (secθ + tanθ) (secθ - tanθ), to get sec²θ - tan²θ.

Now use the identity sec²θ = tan²θ + 1, to simplify it

Reply 8

Abby5001

Original post by Alex621

Once you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)

Expand (secθ + tanθ) (secθ - tanθ), to get sec²θ - tan²θ.

Now use the identity sec²θ = tan²θ + 1, to simplify it

Sorry I wasn't clear, I just meant can you explain how you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)

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Reply 9

Muttley79

Original post by Abby5001

Do you mind explaining in a bit more detail please step by step?

Posted from TSR Mobile

You are trying to get the difference of two squares on the denominator.

It's a bit like rationalising the denominator when you multiply by the conjugate.

Reply 10

Abby5001

Original post by Muttley79

You are trying to get the difference of two squares on the denominator.

It's a bit like rationalising the denominator when you multiply by the conjugate.

Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you

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Reply 11

Alex621

Original post by Abby5001

Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you

Posted from TSR Mobile

I used to hate trig questions in c3 because I never knew where to start but once you've looked at lots of questions you start to get an idea of what you need to do

Reply 12

Abby5001

Original post by Alex621

I used to hate trig questions in c3 because I never knew where to start but once you've looked at lots of questions you start to get an idea of what you need to do

That's good to know, so is multiplying by the conjugate something that is prevalent in c3 because I have never seen it aside from in surds?

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Reply 13

Alex621

Original post by Abby5001

That's good to know, so is multiplying by the conjugate something that is prevalent in c3 because I have never seen it aside from in surds?

Posted from TSR Mobile

I've never seen it before apart from surds but maybe different exam boards use it with trig?

Reply 14

Zacken

Original post by Abby5001

Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you

Posted from TSR Mobile

I think I've done this question recently - the method the markscheme used seem a little convoluted to me, I much prefer what I did!

x = \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta}

and the continue from there converting everything into sines and cosines, collecting and then seperating and converting back to secs. (x + 1/x = 2 sec theta, right?)

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