If tan²A + 2 tan² B+3=0. show that, cos² B + 2 cos²A= 0
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Step-by-step explanation:
tan²A + 2 tan² B+3=0
sin²A/cos²A + 2sin²B/cos²B + 3 = 0
(1 - cos²A)/cos²A + 2[(1 - cos²B)/cos²B] + 3 = 0
(1 - cos²A)cos²B+ 2(1 - cos²B)cos²A + 3 cos²Acos²B = 0
cos²B - cos²Acos²B + 2cos²A - 2cos²Acos²B +3 cos²Acos²B = 0
cos²B + 2cos²A - 3cos²Acos²B+ 3 cos²Acos²B = 0
cos²B + 2cos²A = 0.
Hence proved.
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