
trig identity question
hey
need some help with this one please..
i need to prove that $\displaystyle cot^2 \theta  cos^2 \theta = cot^2 \theta cos^2 \theta$
i took the right hand side and tried to do a similar thing to my text book in proving the question...
this is what I got
$\displaystyle cot^2 \theta cos^2 \theta =$ $\displaystyle (cos^2 \theta  cos^2 \theta.sin^2 \theta) \over(sin^2 \theta)$
$\displaystyle \rightarrow$ $\displaystyle cos^2 \theta\over sin^2\theta$  $\displaystyle cos^2 \theta \dot sin^2\theta \over sin^2 \theta$
$\displaystyle \rightarrow$ RHS = LHS
first of all is this correct? and second of all what rule gives me my first line?
many thanks :)

$\displaystyle cot^2\theta  cos^2\theta = \frac{cos^2\theta}{sin^2\theta}  cos^2\theta$
When you simplify you get the first line.

$\displaystyle cot^2\theta cos^2\theta =$
$\displaystyle = \frac{cos^2\theta cos^2\theta}{sin^2\theta}$ = $\displaystyle \frac{cos^2\theta (1  sin^2\theta)}{sin^2\theta}$