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.

Reply 5

Dysf(x)al

20

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.

Reply 6

Hirsty97

OP

21

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

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?

Reply 8

Hirsty97

OP

21

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

(edited 6 years ago)

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

\displaystyle (\sec\theta + \tan\theta) + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2}{\sec\theta + \tan\theta} + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}

Reply 10

Hirsty97

OP

21

Original post by _gcx

\displaystyle (\sec\theta + \tan\theta) + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2}{\sec\theta + \tan\theta} + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}

Yeah that's it. So I just multiply it by the denominator?

Reply 11

davros

16

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.

C3 AQA Trig identities

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