If tan A + tan B = a and cot A + cot B = b, Prove that cot(A+B) = 1/a − 1/b

Question

If

\[ \tan A+\tan B=a \]

and

\[ \cot A+\cot B=b \]

prove that:

\[ \cot(A+B)=\frac{1}{a}-\frac{1}{b} \]

Proof

Given,

\[ \tan A+\tan B=a \]

\[ \cot A+\cot B=b \]

\[ \frac{1}{\tan A}+\frac{1}{\tan B}=b \]

\[ \frac{\tan A+\tan B}{\tan A\tan B}=b \]

\[ \frac{a}{\tan A\tan B}=b \]

\[ \tan A\tan B=\frac{a}{b} \]

Now,

\[ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B} \]

\[ = \frac{a} {1-\frac{a}{b}} \]

\[ = \frac{ab}{b-a} \]

Therefore,

\[ \cot(A+B) = \frac{b-a}{ab} \]

\[ = \frac{1}{a}-\frac{1}{b} \]

Hence proved.