extremely frustrating identity...

**skoker** · February 22nd 2011, 08:16 AM

this one is giving me headache. I have tried maybe 7 tries and can not get close to what they want.

express the fallowing in terms of a single function.

$sec \theta - tan \theta$ in terms of $sin \theta$ .

I can only use the elementary identities. Pythagorean Reciprocal Co-function even/odd.

the answer to this in the book is $\frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}$ which i can not get.

2 questions...

1) can I take the square root of the expression and multiply through the way you can with an equation? I know you can multiply by the conjugate of the denominator. but im not sure about the other two.

2) I can't find the right strategy. I was thinking get a single term in numerator and denominator that equal part of the Pythagorean that have been squared. and go from there. but that seems to be long way around.

any steps or ideas would be great.

**sa-ri-ga-ma** · February 22nd 2011, 08:40 AM

secθ - tanθ = 1/cosθ - sinθ/cosθ = (1 - sinθ)/cosθ
Now substitute cosθ = sqrt( 1 - sin^2θ) and simplify.

**masters** · February 22nd 2011, 08:49 AM

Originally Posted by skoker

this one is giving me headache. I have tried maybe 7 tries and can not get close to what they want.

express the fallowing in terms of a single function.

$sec \theta - tan \theta$ in terms of $sin \theta$ .

I can only use the elementary identities. Pythagorean Reciprocal Co-function even/odd.

the answer to this in the book is $\frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}$ which i can not get.

2 questions...

1) can I take the square root of the expression and multiply through the way you can with an equation? I know you can multiply by the conjugate of the denominator. but im not sure about the other two.

2) I can't find the right strategy. I was thinking get a single term in numerator and denominator that equal part of the Pythagorean that have been squared. and go from there. but that seems to be long way around.

any steps or ideas would be great.

Hi skoker,

$\sec \theta - \tan \theta = \frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}$

Working from the left side...

[1] $\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}=$

[2] $\frac{1- \sin \theta}{\cos \theta} =$

Using $\cos^2 \theta + \sin^2 \theta=1$ , we show that $\cos \theta=\sqrt{1- \sin^2 \theta}$

[3] $\frac{1- \sin \theta}{\sqrt{1- \sin^2 \theta}}=$

[4] $\frac{1- \sin \theta}{\sqrt{1+\sin \theta} \cdot \sqrt{1-\sin \theta}}\cdot \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}}=$

[5] $\frac{(1-\sin \theta)\sqrt{1-\sin \theta}}{(1-\sin \theta)\sqrt{1+\sin \theta}}=$

[6] $\frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}$

**skoker** · February 22nd 2011, 09:42 AM

AH! thank you masters and sa-ri-ga-ma.

@masters
so you split that denominator into

$\sqrt{1+sin\theta}\sqrt{1-sin\theta}$

because you factor the difference of 2 squares then split the factors?
or because the root is congegate pair of -/+ ?

that was the part i could not get. when i rationalized.

**Soroban** · February 22nd 2011, 12:40 PM

I like sa-ri-ga-ma's suggestion!

$\displaystyle \sec\theta - \tan\theta \;=\;\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} \;=\; \frac{1 - \sin\theta}{\cos\theta}$

$\text{Now substitute }\cos\theta \:=\: \sqrt{1 - \sin^2\!\theta}\,\text{ and simplify.}$

I assume "and simplify" means the following:

$\displaystyle \frac{1-\sin\theta}{\sqrt{1-\sin^2\!\theta}} \;=\;\frac{\left(\sqrt{1-\sin\theta}\right)^2} {\sqrt{(1-\sin\theta)(1+\sin\theta)}}$

. . . . . . . . . $\displaystyle =\;\frac{\sqrt{1-\sin\theta}\,\sqrt{1-\sin\theta}}{\sqrt{1-\sin\theta}\,\sqrt{1+\sin\theta}}$

. . . . . . . . . $\displaystyle =\;\frac{\sqrt{1-\sin\theta}}{\sqrt{1+\sin\theta}}$

**HallsofIvy** · February 23rd 2011, 03:26 AM

Originally Posted by skoker

AH! thank you masters and sa-ri-ga-ma.

@masters
so you split that denominator into

$\sqrt{1+sin\theta}\sqrt{1-sin\theta}$

because you factor the difference of 2 squares then split the factors?

Yes.

or because the root is congegate pair of -/+ ?

Isn't this really the same thing? What do you mean by "conjugate pair"?

that was the part i could not get. when i rationalized.

**skoker** · February 23rd 2011, 02:58 PM

Originally Posted by HallsofIvy

What do you mean by "conjugate pair"?

well this may be hard to explain because I'm not sure its the right terms or ideas but I will explain. I say conjugate pair meaning a pair of pos/neg values. like you have conjugate denominator and complex conjugate. and in the same way you have $\pm a$ where a is a real number. which is the same as $|a|$ which is I think same as $\pm \sqrt{a}$ . but is not the same as $\sqrt[3]{a}$ which is why you can have negative number under cube and only one principle root. little packages of +/- values. now I saw someone 'clear' out an absolute value with this $(\pm a)^2=a$ this sort of struck me. and got me thinking about just what is going on with all this +/- business. and the relation to root and abs and +/-. so I see how this was split into +/- under a root and think for a second could you do that somehow other then factor? some technique I am missing or do not understand involving the fact that there is a square under a conjugate pair root? but then I see that it is a binomial which is more complicated to think about. the books do not go into this. they say you can clear a radical by raising to the power of the index. and this relates to the absolute value. but it seems that the relations between all these ideas are deeper.

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