Question

If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) $\frac{1}{a}-\frac{1}{b}$.

Answer 1

Given:

$$\begin{gathered} \cot A+\cot B=b \\ \Rightarrow \frac{1}{\tan A}+\frac{1}{\tan B}=b \\ \Rightarrow \frac{\tan A+\tan B}{\tan A\tan B}=b \\ \mathrm{Now}, \\ \mathrm{RHS}=\frac{1}{a}-\frac{1}{b} \\ =\frac{1}{\tan A+\tan B}-\frac{\tan A\tan B}{\tan A+tanB} \\ =\frac{1-\tan A\tan B}{\tan A+\tan B} \\ =\cot (A+B) \\ =\mathrm{LHS} \\ \mathrm{Hence}\text{p}\mathrm{roved}. \end{gathered}$$