if A lies in 2nd quadrant and 3tanA +4=0 then solve 2cotA-5cosA + sinA = ?
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3tan A + 4 = 0
3tanA = -4
tanA = -4/3
Assuming a right angled triangle with p=4, b=3
h^2 =p^2 + b^2
h = 5
sinA = 4/5 (+ve because 2nd quadrant)
cosA = -3/5 (-ve because 2nd quadrant)
cotA = -3/4
2cotA - 5cosA + sinA
2*(-3/4) - 5*(-3/5) + 4/5
-1.5 + 3 + 0.8 = 2.3
3tanA = -4
tanA = -4/3
Assuming a right angled triangle with p=4, b=3
h^2 =p^2 + b^2
h = 5
sinA = 4/5 (+ve because 2nd quadrant)
cosA = -3/5 (-ve because 2nd quadrant)
cotA = -3/4
2cotA - 5cosA + sinA
2*(-3/4) - 5*(-3/5) + 4/5
-1.5 + 3 + 0.8 = 2.3
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