$\displaystyle \frac{cos \beta}{1- sin\beta}= sec \beta + tan \beta$
$\displaystyle \frac {1}{sec\beta} / 1- \frac {1}{csc\beta}$
$\displaystyle \frac {1}{1-sec\beta} · \frac{csc\beta}{csc\beta -1}$
Is this right so far?...what do I do next?
$\displaystyle \frac{cos \beta}{1- sin\beta}= sec \beta + tan \beta$
$\displaystyle \frac {1}{sec\beta} / 1- \frac {1}{csc\beta}$
$\displaystyle \frac {1}{1-sec\beta} · \frac{csc\beta}{csc\beta -1}$
Is this right so far?...what do I do next?
$\displaystyle \frac{\cos{b}}{1-\sin{b}} \cdot \frac{1+\sin{b}}{1+\sin{b}} =$
$\displaystyle \frac{\cos{b}(1 + \sin{b})}{1 - \sin^2{b}} =$
$\displaystyle \frac{\cos{b}(1 + \sin{b})}{\cos^2{b}} =$
$\displaystyle \frac{1 + \sin{b}}{\cos{b}} =$
$\displaystyle \frac{1}{\cos{b}} + \frac{\sin{b}}{\cos{b}}$
Hi purplec16,
you can start to reason it through a little also...
$\displaystyle \frac{cos\beta}{1-sin\beta}=sec\beta+tan\beta$
We can write out the definitions for the terms on the right hand side...
$\displaystyle \frac{cos\beta}{1-sin\beta}=\frac{1}{cos\beta}+\frac{sin\beta}{cos\b eta}$
Now we see we need to have $\displaystyle cos\beta$ 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 $\displaystyle \frac{cos\beta}{cos\beta}$
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!
$\displaystyle \frac{cos\beta}{1-sin\beta}\ \frac{cos\beta}{cos\beta}=\frac{cos^2\beta}{cos\be ta(1-sin\beta)}$
Now we are not supposed to have $\displaystyle 1-sin\beta$ under the line
so we need to fish that out!
We therefore multiply by 1 again, this time using the "conjugate" of $\displaystyle 1-sin\beta$
because $\displaystyle (1-sin\beta)(1+sin\beta)=1(1+sin\beta)-sin\beta(1+sin\beta)=1+sin\beta-sin\beta-sin^2\beta$
$\displaystyle =1-sin^2\beta$ which is $\displaystyle cos^2\beta$
Therefore
$\displaystyle \frac{cos^2\beta}{cos\beta(1-sin\beta)}\ \frac{1+sin\beta}{1+sin\beta}=\frac{cos^2\beta(1+s in\beta)}{cos^3\beta}$
$\displaystyle =\frac{1+sin\beta}{cos\beta}=\frac{1}{cos\beta}+\f rac{sin\beta}{cos\beta}$