Math Help - trigonometric identities

1. trigonometric identities

and i dont understand this at all!

can you explain this one?

$\frac{1+sec(-x)}{sin(-x)+tan(-x)}$

2. Originally Posted by singleton2787
and i dont understand this at all!

can you explain this one?

$\frac{1+sec(-x)}{sin(-x)+tan(-x)}$
This problems requires that you know about even/odd functions.

A function is even if $f(-x)=f(x)$, and a function is odd if $f(-x)=-f(x)$.

Let's apply this to trig:

$cos(-x)=cos(x)$ implies that cosine is an even function.
$sin(-x)=-sin(x)$ implies that sine is an odd function.

Thus, we can rewrite our expression as:
$\frac{1+sec(-x)}{sin(-x)+tan(-x)}\implies \frac{1+\frac{1}{cos(-x)}}{sin(-x)+\frac{sin(-x)}{cos(-x)}}\implies =\frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$
$\implies \frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}=-\frac{1}{sin(x)}=-csc(x) \ \forall \ x\neq \frac{k\pi}{2} \ where \ k \ is \ odd$

Hope this helped you out!

3. question

alright thanks that helped a lot
but i dont understand how you got from here....

$\frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$

to here...

$\frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}$

how did you get $1+\frac{1}{cos(x)}$ in the bottom of the second fraction?

4. Originally Posted by singleton2787
alright thanks that helped a lot
but i dont understand how you got from here....

$\frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$

to here...

$\frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}$

how did you get $1+\frac{1}{cos(x)}$ in the bottom of the second fraction?
Pull out a common factor of $-sin(x)$

Thus, $-sin(x)-\frac{sin(x)}{cos(x)}=-sin(x)\left(1+\frac{1}{cos(x)}\right)$

$\therefore \frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}} \implies \frac{1+\frac{1}{cos(x)}}{-sin(x)\left(1+\frac{1}{cos(x)}\right)}\implies \frac{1}{-sin(x)}\cdot \frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x)}}$