If sinA .tanA/cosA=9 than prove that
SinA=3/√10
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sinA.tanA/cosA = 9
{sinA/cosA}.tanA = 9
tanA.tanA = 9
tan² A = 9
take square both sides,
tanA = ± 3
here we take tanA = 3 = 3/1 = p/b
we know, sinA = p/h
h = √(p² + b²) = √(3² + 1²) = √10
hence , sinA = p/h = 3/√10
{sinA/cosA}.tanA = 9
tanA.tanA = 9
tan² A = 9
take square both sides,
tanA = ± 3
here we take tanA = 3 = 3/1 = p/b
we know, sinA = p/h
h = √(p² + b²) = √(3² + 1²) = √10
hence , sinA = p/h = 3/√10
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sins.tanA/CosA=9. sinA.sins/cosA.cosA=9. tanA=3. tan=perpendicular/base. perpendicular=3. base=1. ,hypotenuse=√10 than. sinA=perpendicular/hypotenuse. sinA=3/√10
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