Solution: or, 1/secA-tanA + 1/ secA+tanA = 1/cosA + 1/cosA or, (secA+tanA + secA-tanA)/(secA-tanA )(secA+tanA)= secA + secA Or, 2secA/(sec२A-tan२A)=2secA Therefore, 2secA=2secA proved
Category Archives: Mathematics
If A+B+C=π then,Prove that: cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1
Solution: Here,A + B + C = π then A+B=π-C B+C=π-A A+C=π-C Now, Let’s consider 2 cosA . cosB . cosC = cosA [2 cosB.cosC] =cosA[cos(B+C)+ cos(B- C)] =cosA [cos(180-A) + cos(B-C)] =cosA [-cosA + cos(B-C)] = -cos²A + cosA cos(B-C) = -cos²A + {2cosA cos(B-C)}/2 =-cos²A + {cos(A-B+C) + cos(A+B-C)}/ 2 = -cos²A +Continue reading “If A+B+C=π then,Prove that: cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1”
If A + B + C = π then Prove that: cos²A + cos²B – cos²C+ 2sinA.sinB.cosC=1
Solution: Here,A + B + C = π then B+C=π-A Now, Let’s consider 2 sinA . sinB . cosC = sinA [2 sinB.cosC] = sinA [sin(B+C) + sin(B-C)] =sinA[sin(180-A) + sin(B-C)] = sinA [sinA + sin(B-C)] = sin²A + sinA sin(B-C) = sin²A + {2sinA sin(B-C) }/ 2 = sin²A + {cos(A-B+C) – cos(A+B-C) }/Continue reading “If A + B + C = π then Prove that: cos²A + cos²B – cos²C+ 2sinA.sinB.cosC=1”
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