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If tan A + tanB = a and cot A + cotB = b, prove that: cot(A + b) = 1 a − 1 b . |
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Answer» here i pasted pic alsoso that it can be easy for you to understandStep-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). |
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