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Therefore Dr Peterson's suggestion is a very good hint. Possibly it is not the correct approximation but it gets you to think to approximate.

Alternatively you can think limits. As B gets small, then 1+B approaches 1. Therefore tan(A) approaches 1 and A approaches pi/4.

If A is close to pi/4, then since sec(pi/4)= sqrt(2) it should follow that sec(A) approaches sqrt(2).

Then sqrt(2)(1 + pi/8)= 1.969573929642890829678673847967284790896499586548909351062.

My concern is that sqrt(2)(1 + pi/8) - sec(pi/4) = 0.555360367269795780876985123757586712326827711171961277885 which is not very small! I now have doubts about the statement of your problem.

If in fact my work is incorrect can someone please point out my error. Thanks!

Jomo

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)

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Sweet. That is so clear!Jomo

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)

So what is wrong in my work? It seems to show that the desired result is not correct

When you say the problem is misstated JeffM do you mean it has been phrased badly or I have transcribed the question incorrectly?

I have checked the book, what I wrote is the question asked.

Also I follow you suggestion until how sec is approx. 2. How then root 2 =root 2(1+0/2)?

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You just need to know two things here. The first is that sqrt(2) * 1 = sqrt(2). Then you need to realize that (1+0/2) =1. Putting these two together shows that sqrt(2) = sqrt(2)(1 + 0/2). Where did you get lost?How then root 2 =root 2(1+0/2)?

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You were told what format the equation must be in.Why it can't be left as sec alpha approx. root2? putting the (1+beta/2) back in

What I meant is that when you intended to substitute beta for epsilon in the final expression, you substituted alpha instead. That results in much weirdness.

When you say the problem is misstated JeffM do you mean it has been phrased badly or I have transcribed the question incorrectly?

I have checked the book, what I wrote is the question asked.

Also I follow you suggestion until how sec is approx. 2. How then root 2 =root 2(1+0/2)?

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The problem statement should be:Dear All

This question appears in a British textbook aimed at 16 year olds, at the end of a chapter on the "The General Angle".

View attachment 16629

I am struggling!. I have tried using sec ^2 =1 +tan^2 identity to no avail, also not sure where beta is small comes into it.

A few clues to start would be appreciated.

If

tan(a) = 1 + b, ............where b is small,

prove that

sec(a) = \(\displaystyle \pm \sqrt{2}\) (1 + b/2)

sec

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An important thing we need to know in order to help you toward the solution the isThis question appears in a British textbook aimed at 16 year olds, at the end of a chapter on the "The General Angle".

View attachment 16629

I am struggling!. I have tried using sec ^2 =1 +tan^2 identity to no avail, also not sure where beta is small comes into it.

I'm very curious about what was taught!

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Thank you for clarifying that!The problem statement should be:

If

tan(a) = 1 + b, ............where b is small,

prove that

sec(a) = \(\displaystyle \pm \sqrt{2}\) (1 + b/2)

sec^{2}(a) = 1 + tan^{2}(a) = 1 + (1 + b)^{2}= 2 + 2b +~~b~~^{2}= 2 * (1 + b) .......continue.....

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Oh, now I see your point. The OP put alpha for beta.

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)