and i dont understand this at all!(Thinking)

can you explain this one?

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

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- May 19th 2008, 08:12 PMsingleton2787trigonometric identities
and i dont understand this at all!(Thinking)

can you explain this one?

$\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$ - May 19th 2008, 08:21 PMChris L T521
This problems requires that you know about even/odd functions.

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

Let's apply this to trig:

$\displaystyle cos(-x)=cos(x)$ implies that cosine is an even function.

$\displaystyle sin(-x)=-sin(x)$ implies that sine is an odd function.

Thus, we can rewrite our expression as:

$\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}\implies \frac{1+\frac{1}{cos(-x)}}{sin(-x)+\frac{sin(-x)}{cos(-x)}}\implies =\frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$

$\displaystyle \implies \frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}=-\frac{1}{sin(x)}=-csc(x) \ \forall \ x\neq \frac{k\pi}{2} \ where \ k \ is \ odd$

Hope this helped you out! :D - May 19th 2008, 08:32 PMsingleton2787question
alright thanks that helped a lot

but i dont understand how you got from here....

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

to here...

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

how did you get $\displaystyle 1+\frac{1}{cos(x)}$ in the bottom of the second fraction? - May 19th 2008, 09:25 PMChris L T521
Pull out a common factor of $\displaystyle -sin(x)$

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

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