Thanks!

2. (i) $\displaystyle sin(x)cos(x)sec(x)cot(x) = sin(x)cos(x)\frac{1}{cos(x)}\frac{cos(x)}{sin(x)} = cos(x)$
(ii) $\displaystyle (1+cos(x))(cosec(x)-cot(x)) = (1+cos(x))(\frac{1}{sin(x)}-\frac{cos(x)}{sin(x)})$$\displaystyle = (1+cos(x))(\frac{1-cos(x)}{sin(x)})$
$\displaystyle = \frac{(1+cos(x))(1-cos(x)}{sin(x)}=\frac{1-cos^2(x)}{sin(x)} =\frac{sin^2(x)}{sin(x)} = sin(x)$
(iii)$\displaystyle \frac{sin(x)}{1+cos(x)}+\frac{sin(x)}{1-cos(x)} = \frac{sin(x)(1-cos(x))}{(1+cos(x))(1-cos(x))}+\frac{sin(x)(1+cos(x))}{(1+cos(x))(1-cos(x))}$
$\displaystyle \frac{sin(x)-sin(x)cos(x)+sin(x)+sin(x)cos(x)}{1-cos^2(x)}= \frac{2sin(x)}{sin^2(x)} =\frac{2}{sin(x)} = 2 csc(x)$

3. wow, thanks a lot for the quick reply, i would have had that right except for a silly mistake!

are you, or anyone else for that matter, able to help out with the last 2?

thanks!

EDIT: WOW! THAT WAS FAST! You posted another answer while I was typing the reply! Thanks a lot!

4. Hello, gobbajeezalus!

Do you know any of the basic identities?

$\displaystyle 1)\;\;(\sin x)(\cos x)(\sec x)(\cot x)$

We have: .$\displaystyle ({\color{red}\rlap{/////}}\sin x)\cdot(\cos x)\cdot\left(\frac{1}{{\color{green}\rlap{/////}}\cos x}\right)\cdot\left(\frac{{\color{green}\rlap{/////}}\cos x}{{\color{red}\rlap{/////}}\sin x}\right) \;=\;\cos x$

$\displaystyle 2)\;\;(1 + \cos x)(\csc x - \cot x)$

We have: .$\displaystyle \left(\frac{1 + \cos x}{1}\right)\cdot\left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) \;=$ .$\displaystyle \frac{1 + \cos x}{1}\cdot\frac{1-\cos x}{\sin x}$

. . . . . . $\displaystyle = \;\frac{1-\cos^2x}{\sin x} \;\;=\;\;\frac{\sin^2\!x}{\sin x} \;\;=\;\;\sin x$

$\displaystyle 3)\;\;\frac{\sin x}{1+\cos x} + \frac{\sin x}{1-\cos x}$
Get a common denominator . . .

$\displaystyle \frac{\sin x}{1+\cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;+\; \frac{\sin x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;\;=\;\;\frac{\sin x(1 - \cos x)}{1-\cos^2\!x} \;+\; \frac{\sin x(1+\cos x)}{1-\cos^2\!x}$

. . $\displaystyle = \;\;\frac{\sin x(1-\cos x)}{\sin^2\!x} + \frac{\sin x(1+\cos x)}{\sin^2\!x} \;\;=\;\;\frac{1-\cos x}{\sin x} + \frac{1+\cos x}{\sin x}$ . $\displaystyle =\;\;\frac{2}{\sin x} \;\;=\;\;2\csc x$