Question

If tan A + tanB = a and cot A + cotB = b, prove
that: cot(A + b) =
1
a
−
1
b
.

Answer 1

Answer:

here i pasted pic also

so that it can be easy for you to understand

Step-by-step explanation:

If tanA+tanB = a, then sinA/cosA + sinB/cosB = a, multiply both sides by cosAcosB gets you sinAcosB + sinBcosA = acosAcosB. Substituting sin(A+B) = sinAcosB+sinBcosA and cosAcosB = (1/2)cos(A+B) + (1/2)cos(A-B) yields sin(A+B) = (1/2)*a*[cos(A+B) + cos(A-B)]. Therefore, 1/a = (1/2)*cot(A+B) + (1/2)*cos(A-B)/sin(A+B).

Similarly, cotA+cotB = b, gets cosA/sinA + cosB/sinB = b and cosAsinB + sinAcosB = bsinAsinB. Substituting yields sin(A+B) = (-1/2)*b[cos(A+B) - cos(A-B)]. Therefore 1/b = (-1/2)*cot(A+B) + (1/2)*cos(A-B)/sin(A+B).

When you subtract 1/b from 1/a, the second terms on the right hand side cancel out, and you are left with 1/a - 1/b = cot(A+B).