please answer to the 17th and 12th question in the photo
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1+sinA.sinA = 3sinA.cosA divide by sinA.cosA
(sin^2 A + cos^2 A) / (sinA.cosA ) + tan A = 3
sin^2 A / (sinA.cosA ) + cos^2 A / (sinA.cosA ) + tan A = 3
tan A + 1/tan A + tan A = 3
2tan^2 A - 3tan A +1 = 0
(2tan A -1)(tan A -1) = 0
then
tan A = 1 or tan A = 1/2
Hope it helps.
Pl. mark as brainliest.
(sin^2 A + cos^2 A) / (sinA.cosA ) + tan A = 3
sin^2 A / (sinA.cosA ) + cos^2 A / (sinA.cosA ) + tan A = 3
tan A + 1/tan A + tan A = 3
2tan^2 A - 3tan A +1 = 0
(2tan A -1)(tan A -1) = 0
then
tan A = 1 or tan A = 1/2
Hope it helps.
Pl. mark as brainliest.
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