and i dont understand this at all!
can you explain this one?
$\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$
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!
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?
Pull out a common factor of $\displaystyle -sin(x)$
Thus, $\displaystyle -sin(x)-\frac{sin(x)}{cos(x)}=-sin(x)\left(1+\frac{1}{cos(x)}\right)$
$\displaystyle \therefore \frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}} \implies \frac{1+\frac{1}{cos(x)}}{-sin(x)\left(1+\frac{1}{cos(x)}\right)}\implies \frac{1}{-sin(x)}\cdot \frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x)}}$