If tan A + tan B = p and cot A + cot B = q, then cot (A + B) is:

Correct option is B

Given:

tan A + tan B = p

cot A + cot B = q 

Formula Used:

​$$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$ 

Solution:

tan A + tan B = p

ot A + cot B = q 

​$$\frac{1}{\tan A} + \frac{1}{\tan B} = q$$

$$\frac{\tan B + \tan A}{\tan A \cdot \tan B} = q$$ 

$$\frac{p}{\tan A \cdot \tan B} = q$$

tanA.tanB =p\q_____(eq2)

as we know 

​$$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$ 

$$\tan(a + b) = \frac{p}{1 - \tan(a)\tan(b)}$$​​

put the value of tanA.tanB =p\q 

​$$\tan(a + b) = \frac{p}{1 - \frac{p}{q}}$$ 

tan(a+b)=$$\frac{pq}{q-p}$$ 

cot(a+b) =$$\frac{q-p}{pq}$$​​