Originally Posted by
Moo Hello,
Please put parenthesis...
I guess the second one is :
$\displaystyle \frac{\sin^2 x}{1- \cos x}$
We know that $\displaystyle \sin^2 x+\cos^2 x=1 \implies \sin^2 x=1-\cos^2 x=(1-\cos x) \cdot (1+\cos x)$
So, how does it simplify ?
(not sure about this one)
$\displaystyle \frac{\sec^2 x-1}{\sec^2 x}$ (or $\displaystyle \sec^2 x-\frac{1}{\sec^2 x}$ ?)
$\displaystyle =1-\frac{1}{\sec^2 x}=1-\cos^2 x=...$
Is the first one $\displaystyle \frac{\sin x}{1+\cos x}+\cot x$ or $\displaystyle \frac{\sin x}{1+\cos x+\cot x}$ ?