Question

If $A+B+C=\pi /2$, then show that
i. $\cot A+\cot B+\cot C=\cot A\cot B\cot C$
ii. $\tan A\tan B+\tan B\tan C+\tan C\tan A=1$.

Answer 1

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

$\cot A+\cot B+\cot C$

simplifying in terms of sin and cos

$=\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}$

$=\frac{\left(\cos A\sin B\sin c+\cos B\sin A\sin c+\cos C\sin A\sin B\right)}{\sin A\sin B\sin C}$

$=\frac{-\cos (A+B+C)+\cos A\cos B\cos C}{\sin A\sin B\sin C}=\frac{\cos A\cos B\cos C}{\sin a\sin B\sin C}=\cot A\cot B\cot C$
$\tan A\tan B+\tan B\tan C+\tan C\tan A$

simplifying in terms of sin and cos

$=\frac{\sin A}{\cos A}.\frac{\sin B}{\cos B}.\frac{\sin C}{\cos C}+\frac{\sin C}{\cos C}.\frac{\sin A}{\cos A}$

$=\frac{\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B}{\cos A\cos B\cos C}$

$=\frac{\cos A\cos B\cos C-\cos (A+B+C)}{\cos A\cos B\cos C}=1$