Question

Prove that in $\Delta ABC$: $cosecA.cosecB.cosecC+\cot A\cot B\cot C=\cot A+\cot B+\cot C$.

Answer 1

Given: $A+B+C+180$........As it is a triangle

We know that $\cos (A+B+C)=\cos {180}^{\circ }=-1$

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

$\therefore 1+\cos A\cos B\cos C=\cos A\sin B\sin C+\cos B\sin C\sin A+\cos C\sin A\sin B$

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

$\therefore cosecA.cosecB.cosecC+\cot A\cot B\cot C=\cot A+\cot B+\cot C$.