# Prove for trianlge ABC

• Feb 24th 2010, 07:20 PM
arze
Prove for trianlge ABC
1. $\cot C=\frac{a}{c}\csc B-\cot B$
I have not idea where to start here.

2. $abc=4\bigtriangleup R$ where R is the radius of the circumcircle, and $\bigtriangleup$ is the area of the triangle.
I used $\bigtriangleup=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{1}{2}(a+b+c)$ and got the value for $\bigtriangleup=\frac{1}{4}\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2}$ which doesn't have a square root. Now I'm stuck
Thanks
• Feb 25th 2010, 01:44 AM
Hello arze
Quote:

Originally Posted by arze
1. $\cot C=\frac{a}{c}\csc B-\cot B$
I have not idea where to start here.

2. $abc=4\bigtriangleup R$ where R is the radius of the circumcircle, and $\bigtriangleup$ is the area of the triangle.
I used $\bigtriangleup=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{1}{2}(a+b+c)$ and got the value for $\bigtriangleup=\frac{1}{4}\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2}$ which doesn't have a square root. Now I'm stuck
Thanks

1. Use the Sine Rule on the RHS, by writing
$\frac{a}{c}=\frac{\sin A}{\sin C}$
and write $\csc B$ and $\cot B$ in terms of $\sin B$ and $\cos B$.

So:
$\frac{a}{c}\csc B-\cot B =\frac{\sin A}{\sin C\sin B}-\frac{\cos B}{\sin B}$
$=\frac{\sin A-\sin C\cos B}{\sin C\sin B}$
Now use the fact that $\sin A = \sin (180^o-A) = \sin(B+C)$. Expand this, and then simplify the numerator. Can you finish this now?

2. Use $\Delta = \tfrac12bc\sin A$, and $\frac{a}{\sin A}=2R$ and you're done.