If $\tan\alpha, \tan\beta$ are the roots of the equation $x^{2}+px+q=0(p\neq 0)$, then
- $\displaystyle \tan(\alpha+\beta)=\frac{1-q}{p}$
- $\displaystyle \tan(\alpha+\beta)=\frac{-p}{q-1}$
- $\displaystyle \tan(\alpha+\beta)=\frac{p}{q-1}$
- $\cot(\alpha+\beta)=p$
If tan(alpha) and tan(beta) are roots, tan(alpha) + tan(beta) = -p and tan(alpha) * tan(beta) = q. Using the identity tan(alpha + beta) = (tan(alpha) + tan(beta)) / (1 - tan(alpha) * tan(beta)), we get (-p) / (1 - q) = p / (q - 1).
Using Vieta's formulas for the quadratic equation x^2 + px + q = 0, the sum of the roots tan alpha and tan beta equals -p, and their product equals q. According to the tangent addition formula, tan(alpha + beta) = (tan alpha + tan beta) / (1 - tan alpha tan beta). Substituting the values from Vieta's formulas into this identity gives tan(alpha + beta) = -p / (1 - q). By multiplying the numerator and denominator by -1, this simplifies exactly to p / (q - 1), making this the correct expression.