Question

If A, B, C are the angles of a triangle, show that $\frac{\cot A+\cot B}{\tan A+\tan B}+\frac{\cot B+\cot C}{\tan B+\tan C}+\frac{\cot C+\cot A}{\tan C+\tan A}=1$.

Answer 1

Convert $\cot$ in terms of $\tan$ in the left hand side of the given equation.

$\frac{\tan A+\tan B}{\tan A\tan B\left(\tan A+\tan B\right)}+\frac{\tan B+\tan C}{\tan B\tan C\left(\tan B+\tan C\right)}+\frac{\tan C+\tan A}{\tan C\tan A\left(\tan C+\tan A\right)}=1$

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

$\tan C+\tan A+\tan B=\tan A\tan B\tan C$

Therefore, by the sum of angles formula for tangent, $LHS=RHS$.