Question #b9c5a

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Question #b9c5a

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Answer 1 · 2016-08-15T10:27:48

Given
$cos e c A = \frac{2}{\sqrt{3}}$ $cosecA=2/sqrt3$
$\Rightarrow cos e c A = cos e c (\frac{π}{3}) = cos e c (π - \frac{π}{3})$ $=>cosecA=cosec(pi/3)=cosec(pi-pi/3)$
(Takling $2 π > A > 0$ $2pi>A>0$ )

$So A = \frac{π}{3} or \frac{2 π}{3}$ $"So "A =pi/3 or (2pi)/3$

So $tan A = \frac{tan π}{3} = \sqrt{3}$ $tanA=tanpi/3=sqrt3$

or $tan A = \frac{tan (2 π)}{3} = tan (π - \frac{π}{3})$ $tanA=tan(2pi)/3=tan(pi-pi/3)$
$= - \frac{tan π}{3} = - \sqrt{3}$ $=-tanpi/3=-sqrt3$

Alternative

Given
$cos e c A = \frac{2}{\sqrt{3}}$ $cosecA=2/sqrt3$

$\Rightarrow {(cos e c A)}^{2} = {(\frac{2}{\sqrt{3}})}^{2}$ $=>(cosecA)^2=(2/sqrt3)^2$

$\Rightarrow cos e c^{2} A = \frac{4}{3}$ $=>cosec^2A=4/3$

$\Rightarrow 1 + {cot}^{2} A = \frac{4}{3}$ $=>1+cot^2A=4/3$

$\Rightarrow {cot}^{2} A = \frac{4}{3} - 1 = \frac{1}{3}$ $=>cot^2A=4/3-1=1/3$

$\Rightarrow {tan}^{2} A = 3$ $=>tan^2A=3$

$\Rightarrow tan A = \pm \sqrt{3}$ $=>tanA=+-sqrt3$

Given
$cos e c A = + v e \to$ $cosecA=+ve->$ A lying in 1st or 2nd quadrant.
When A is in 1st quadrant then $tan A = \sqrt{3}$ $tanA =sqrt3$

When A is in 2nd quadrant then $tan A = - \sqrt{3}$ $tanA =-sqrt3$

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Question #b9c5a

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