Question

If $A+B+C=\pi$, prove that, $\cot A+\cot B+\cot C=\cot A\cot B\cot C+\csc A\csc B\csc C$.

Answer 1

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

$\cos \left(A+B+C\right)=\cos \pi$

$\cos \left(A+B+C\right)=-1$

$\cos \left(A+B\right)\cos C-\sin \left(A+B\right)\sin C=-1$

$\left(\cos A\cos B-\sin A\sin B\right)\cos C-\left(\sin A\cos B+\cos A\sin B\right)\sin C=-1$

$\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C=-1$

Divide both sides by $\sin A\sin B\sin C$,

$\frac{\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C}{\sin A\sin B\sin C}=\frac{-1}{\sin A\sin B\sin C}$

$\cot A\cot B\cot C-\cot C-\cot B-\cot A=-\csc A\csc B\csc C$

$\cot A\cot B\cot C+\csc A\csc B\csc C=\cot A+\cot B+\cot C$

Hence proved.