Question

If TanA+TanB=p and CotA+CotB=q
then Cot(A+B)=

Answer 1

Answer:-

Given:

tan A + tan B = p -- equation (1)

cot A + cot B = q -- equation (2)

From equation (2) , using cot A = 1/tan A and cot B = 1/tan B in LHS we get,

→ (1/tan A + 1/tan B) = q

→ (tan B + tan A) / tan A tan B = q

→ (tan A + tan B) * 1/(tan A * tan B) = q

→ cot A * cot B = q / tan A + tan B

Putting the value of "tan A + tan B" as p we get,

→ cot A * cot B = q/p

We know that,

cot (A + B) = (Cot A * cot B - 1) / (cot A + cot B)

Putting the values of cot A * cot B and cot A + cot B we get,

→ cot (A + B) = (q/p - 1) / (q)

→ Cot (A + B) = [ (q - p) / p ] / q

→ Cot (A + B) = (q - p) / p * 1/q

→ Cot(A + B) = (q - p) / pq

Answer 2

Answer:

the answer is cot (a+b)= (p-q)/p