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# C3 AQA Trig identities watch

1. Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
2. (Original post by Hirsty97)
Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
What have you tried? Please post your working.
3. I'd start by writing as a single fraction. Remember that and . This'll make evaluating slightly nicer.
4. (Original post by Notnek)
What have you tried? Please post your working.
I don't know. I get sec ϴ + tan ϴ + 1/sec ϴ + tan ϴ. I figure I have to get it all as a single fraction
5. 1/X is 1/sec(theta)+tan(theta). Multiply top and bottom by sectheta-tan theta and you will get sectheta-tantheta over sec^2theta-tan^2theta. And we know from the identity of tan^2+1=sec^2 that sec^2-tan^2=1. You simply substitute it in simplify and get your answer.
6. (Original post by Hirsty97)
Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
Try expressing (secϴ + tanϴ) as a single fraction. You know that secϴ = 1/cosϴand that tanϴ= sinϴ/cosϴ. Both of these have the same denominator, so you can group them together. To find 1/x, all you have to do is turn the whole fraction upside down.
7. (Original post by Radioactivedecay)
1/X is 1/sec(theta)+tan(theta). Multiply top and bottom by sectheta-tan theta and you will get sectheta-tantheta over sec^2theta-tan^2theta. And we know from the identity of tan^2+1=sec^2 that sec^2-tan^2=1. You simply substitute it in simplify and get your answer.
Textbook says first step is:

(Sec ϴ + tan ϴ)^2 + 1 / sec ϴ + tan ϴ

I get what to do then but the answer is apparently 2 sec ϴ
8. (Original post by Hirsty97)
Textbook says first step is:

(Sec ϴ + tan ϴ)^2 + 1 / sec ϴ + tan ϴ

I get what to do then but the answer is apparently 2 sec ϴ
What did you do from there?
9. (Original post by _gcx)
What did you do from there?
Expand the brackets

Then sec^2 ϴ = tan^2ϴ + 1

So you get sec^2ϴ + 2secϴtanϴ + sec^2ϴ \ secϴ + tanϴ

Which you can factories and cancel out to be left with 2 sec ϴ

I get that just not the first part to go from:
(Sec ϴ + tan ϴ) + 1/ sec ϴ + tan ϴ

To:
(Sec ϴ + tan ϴ) ^2 + 1/ sec ϴ + tan ϴ
10. (Original post by Hirsty97)
Expand the brackets

Then sec^2 ϴ = tan^2ϴ + 1

So you get sec^2ϴ + 2secϴtanϴ + sec^2ϴ \ secϴ + tanϴ

Which you can factories and cancel out to be left with 2 sec ϴ

I get that just not the first part to go from:
Sec ϴ + tan ϴ + 1/ sec ϴ + tan ϴ

To:
(Sec ϴ + tan ϴ) ^2 + 1/ sec ϴ + tan ϴ
11. (Original post by _gcx)
Yeah that's it. So I just multiply it by the denominator?
12. (Original post by Hirsty97)
Yeah that's it. So I just multiply it by the denominator?
I wouldn't multiply by anything at this stage.

Expand the square bracket on the numerator and use a standard trig formula that relates tan^2 to sec^2.

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