# trig identity question

• Aug 12th 2010, 08:53 PM
stabza
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 :)
• Aug 12th 2010, 09:18 PM
sa-ri-ga-ma
$\displaystyle cot^2\theta - cos^2\theta = \frac{cos^2\theta}{sin^2\theta} - cos^2\theta$

When you simplify you get the first line.
• Aug 13th 2010, 05:30 PM
grgrsanjay
$\displaystyle cot^2\theta cos^2\theta =$
$\displaystyle = \frac{cos^2\theta cos^2\theta}{sin^2\theta}$ = $\displaystyle \frac{cos^2\theta (1 - sin^2\theta)}{sin^2\theta}$