# Simple Trig Help #2

• Dec 13th 2009, 03:04 AM
jgv115
Simple Trig Help #2
Can someone check this:

$(\cos60-\tan45)(\sin30+\tan45)$

$(\frac{\sqrt3}{2} - 1)(\frac {\sqrt3}{2} +1)$

$\frac{3}{4} -1$

$\frac {-3}{4}$

• Dec 13th 2009, 03:06 AM
Prove It
Quote:

Originally Posted by jgv115
Can someone check this:

$(\cos60-\tan45)(\sin30+\tan45)$

$(\frac{\sqrt3}{2} - 1)(\frac {\sqrt3}{2} +1)$

$\frac{3}{4} -1$

$\frac {-3}{4}$

$\sin{30^{\circ}} = \frac{1}{2}$ and $\cos{60^{\circ}} = \frac{1}{2}$

So $(\cos{60^\circ} - \tan{45^\circ})(\sin{30^\circ} + \tan{45^\circ}) = \left(\frac{1}{2} - 1\right)\left(\frac{1}{2} + 1\right)$

$= \left(-\frac{1}{2}\right)\left(\frac{3}{2}\right)$

$= -\frac{3}{4}$.
• Dec 14th 2009, 03:37 AM
jgv115
Simplify

$\frac {2\csc^2 \theta-2}{4\cot\theta}$

Well I just factorised the top bit:

$\frac {2(\csc^2\theta-1)}{4\cot\theta}$

I don't know where to go from here. Any help will be appreciated.
• Dec 14th 2009, 04:59 AM
Prove It
Quote:

Originally Posted by jgv115
Simplify

$\frac {2\csc^2 \theta-2}{4\cot\theta}$

Well I just factorised the top bit:

$\frac {2(\csc^2\theta-1)}{4\cot\theta}$

I don't know where to go from here. Any help will be appreciated.

$\sin^2{\theta} + \cos^2{\theta} = 1$.
Divide both sides by $\sin^2{\theta}$ and you get
$1 + \cot^2{\theta} = \csc^2{\theta}$.
So what do you think $\csc^2{\theta} - 1$ equals?