Identities

**purplec16** · March 7th 2010, 04:00 PM

$\frac{cos \beta}{1- sin\beta}= sec \beta + tan \beta$
$\frac {1}{sec\beta} / 1- \frac {1}{csc\beta}$
$\frac {1}{1-sec\beta} · \frac{csc\beta}{csc\beta -1}$
Is this right so far?...what do I do next?

**skeeter** · March 7th 2010, 04:27 PM

$\frac{\cos{b}}{1-\sin{b}} \cdot \frac{1+\sin{b}}{1+\sin{b}} =$

$\frac{\cos{b}(1 + \sin{b})}{1 - \sin^2{b}} =$

$\frac{\cos{b}(1 + \sin{b})}{\cos^2{b}} =$

$\frac{1 + \sin{b}}{\cos{b}} =$

$\frac{1}{\cos{b}} + \frac{\sin{b}}{\cos{b}}$

**purplec16** · March 7th 2010, 05:05 PM

Thanks so much, how do u you understand how to do these things, when you do it I undestand, but when the question is there on its own I dont know what to do, I guess it calls for a lot of practice

**skeeter** · March 7th 2010, 05:09 PM

Originally Posted by purplec16

Thanks so much, how do u you understand how to do these things, when you do it I undestand, but when the question is there on its own I dont know what to do, I guess it calls for a lot of practice

bingo!

**Archie Meade** · March 7th 2010, 05:39 PM

Hi purplec16,

you can start to reason it through a little also...

$\frac{cos\beta}{1-sin\beta}=sec\beta+tan\beta$

We can write out the definitions for the terms on the right hand side...

$\frac{cos\beta}{1-sin\beta}=\frac{1}{cos\beta}+\frac{sin\beta}{cos\b eta}$

Now we see we need to have $cos\beta$ under the line on the left hand side,

therefore as we can only multiply by 1 (so the ratio stays the same),

we multiply the left hand side by $\frac{cos\beta}{cos\beta}$

as this gets us that term in the denominator,
however we see later on, that this term comes in anyway, so it ends up being unnecessary!

$\frac{cos\beta}{1-sin\beta}\ \frac{cos\beta}{cos\beta}=\frac{cos^2\beta}{cos\be ta(1-sin\beta)}$

Now we are not supposed to have $1-sin\beta$ under the line
so we need to fish that out!

We therefore multiply by 1 again, this time using the "conjugate" of $1-sin\beta$

because $(1-sin\beta)(1+sin\beta)=1(1+sin\beta)-sin\beta(1+sin\beta)=1+sin\beta-sin\beta-sin^2\beta$

$=1-sin^2\beta$ which is $cos^2\beta$

Therefore

$\frac{cos^2\beta}{cos\beta(1-sin\beta)}\ \frac{1+sin\beta}{1+sin\beta}=\frac{cos^2\beta(1+s in\beta)}{cos^3\beta}$

$=\frac{1+sin\beta}{cos\beta}=\frac{1}{cos\beta}+\f rac{sin\beta}{cos\beta}$

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