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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jun 2007 11:46:05 IST
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sec A - tan A = 3 then A lies in which QUADRANT
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a lies in 4th quadrant
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Chandana |
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24 Jun 2007 13:31:34 IST
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objective approach:- secA = 5/3 and tanA = -4/3 satisfies.......... therefore, A lies in fourth quadrant
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Rates are always welcome!
     
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24 Jun 2007 13:43:01 IST
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cos-1(p/a) + cos-1(q/b)=(alpha), then
(p^2/a^2) - 2pq/ab cos(alpha) + q^2/b^2 =?
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Chandana |
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24 Jun 2007 19:26:33 IST
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Given secA - tan A = 1/3 ------------(1)
We know, sec2A - tan2A = 1
(secA + tan A)(secA - tan A) = 1
(secA + tan A) = 3 ----------(2)
Adding (1) and (2)
secA = 5/3
tanA = -4/3
tan is ve and cos (sec) is +ve. Hence A lies in 4th quadrant.
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4 Jul 2007 15:51:24 IST
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secA - tanA =3 Hence secA + tanA =1/3.upon solving the two equations we get secA=5/3 and tanA= -4/3.By this we can say A lies in fourth quardrant
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4 Jul 2007 15:53:56 IST
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A lies in fourth quardrant
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5 Jul 2007 11:07:26 IST
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it lies in fourth quadrant
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