Can someone please prove this equation:
Cos x + Cos x = 2Sec x
1 - Sin x 1 + Sin x
You should consider only one hand side and should end up with 2sec(x).
$\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = \frac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}$
Expand, simplify.
Ultimately, you should get;
$\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = 2sec(x)$
Yes, the answer is $\displaystyle \frac{2}{cos(x)}$ but that is not how you solve a prove question.
There is no point in saying 2/cos x = 2/cos x, because of course that equation is true, and what you have basicly done is rewritten 2sec(x) in the form 2/cos(x).
You should start off on the left, ignoring the right hand side, having only the equation $\displaystyle \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)}$ and simplifying/rearranging it to match what is stated.
You must show the steps of how you got that, without the answer of 2sec(x) in the equation:
$\displaystyle \dfrac{cos(x)}{1-sin(x)} + \dfrac{cos(x)}{1+sin(x)}$
$\displaystyle = \dfrac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}$
$\displaystyle = \dfrac{cos(x) + cos(x)sin(x) + cos(x) - cos(x)sin(x)}{1-sin^2(x)}$
Continue simplifying/rearranging it to get 2sec(x).
Hint: ($\displaystyle cos^2(x) + sin^2(x) = 1$)