# Math Help - Need help with prooving

1. ## Need help with prooving

Im not a very good mathematician, and im only in grade 10..and need some help

Question is: Show
sin(180-sinB)
____________ = -tanB
cos(180-B)

Sorry if this question is really easy, i must really suck.
Ive tried doing it:

sinB
____ = tan B
cosB

.: sin B- sinx
___________ = -tanx
cos B- cosx

Is this method right, and in an exam will i get full marks for it?

Another question:
Simplify cos(90-b)
_________
sin(180-b)

I havent the slightest idea what to do.
I understand that cosx over sinx= 1/2 x tanx..but the values of cos and sin are different in this question... so is it implying that the answer is:

cos90- cos b
_____________ = 1/4 x tan (since cos over sin is 1/2tan, this is half)
sin 180- sin b

And i need to solve for b....

2. Originally Posted by xanith_47
Im not a very good mathematician, and im only in grade 10..and need some help

Question is: Show
sin(180-B)
____________ = -tanB
cos(180-B)
Use the difference formulas for $\cos$ and $\sin$:

$
\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)
$

and:

$
\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)
$

So if $A=180^\circ$:

$
\sin(180-B)=\sin(180)\cos(B)-\cos(180)\sin(B)$
$=\sin(B)
$

$
\cos(180-B)=\cos(180)\cos(B)+\sin(180)\sin(B)$
$=-\cos(B)$

Hence:

$
\frac{\sin(180-B)}{\cos(180-B)}=\frac{\sin(B)}{-\cos(B)}=-tan(B)
$

RonL