If $$\tan A + \tan B = a$$ and $$\cot A + \cot B = b$$, then $$\frac{1}{a} - \frac{1}{b}$$  is equal to .......

$$\tan A + \tan B = a$$  ---(1)

$$\cot A + \cot B = b$$ ---(2)

From equation (1), $$\frac{1}{tan A}$$ + $$\frac{1}{tan B}$$ = a

Or $$\frac{cot A + cot B}{cot A cotB}$$ = a

Or $$\frac{B}{cot A cot B}$$=a

Or $$\frac{b}{a}$$= cot A cot B

Now, $$\color{blue}\cot (A + B)$$= $$\color{blue}\frac{cot A cot B - 1 }{cot A + cot B}$$

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

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

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

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