Question

Prove that if $A+B+C=\pi ,\cot \frac{A}{32}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}$

Answer 1

As $A+B+C=\pi$, so,

$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi }{2}$

$\frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}$

Multiply both sides by tan,

$\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)$

$\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\cot \frac{C}{2}$

$\frac{\frac{1}{\cot \frac{A}{2}}+\frac{1}{\cot \frac{B}{2}}}{1-\left(\frac{1}{\cot \frac{A}{2}}\right)\left(\frac{1}{\cot \frac{B}{2}}\right)}=\cot \frac{C}{2}$

$\frac{\cot \frac{B}{2}+\cot \frac{A}{2}}{\cot \frac{A}{2}\cot \frac{B}{2}-1}=\cot \frac{C}{2}$

$\cot \frac{B}{2}+\cot \frac{A}{2}=\left(\cot \frac{A}{2}\cot \frac{B}{2}-1\right)\cot \frac{C}{2}$

$\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$