if sec A + tan A = p, prove that sin A=p2-1/p2+1
Asked by devanshmudgal
| 12th Sep, 2010,
06:46: PM
Expert Answer:
sec A + tan A = p
(1 + sinA)/cosA = p
(1 + sinA) = pcosA
Square on both sides,
1 + 2sinA + sin2A = p2cos2A
1 + 2sinA + sin2A = p2(1 - sin2A)
1 - p2 + 2sinA + (1 + p2)sin2A = 0
This is a quadratic equation in sinA, the roots are,
sinA = {-2 ± [4 - 4(1 - p2)(1 + p2)]1/2}/ 2(1 + p2)
sinA = {-2 ± [4 - 4(1 - p4)]1/2}/ 2(1 + p2)
sinA = {-2 ± [4p4]1/2}/ 2(1 + p2)
sinA = {-2 ± 2p2}/ 2(1 + p2)
sinA = {p2 - 1}/(1 + p2) .. taking the positive root.
Regards,
Team,
TopperLearning.
(1 + sinA) = pcosA
Square on both sides,
1 + 2sinA + sin2A = p2cos2A
1 + 2sinA + sin2A = p2(1 - sin2A)
1 - p2 + 2sinA + (1 + p2)sin2A = 0
This is a quadratic equation in sinA, the roots are,
sinA = {-2 ± [4 - 4(1 - p2)(1 + p2)]1/2}/ 2(1 + p2)
sinA = {-2 ± [4 - 4(1 - p4)]1/2}/ 2(1 + p2)
sinA = {-2 ± [4p4]1/2}/ 2(1 + p2)
sinA = {-2 ± 2p2}/ 2(1 + p2)
sinA = {p2 - 1}/(1 + p2) .. taking the positive root.
Regards,
Team,
TopperLearning.
Answered by
| 12th Sep, 2010,
09:22: PM
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