Question

If sec2A=3 , where A lies in first quadrant , find sin A

Answer 1

Answer:

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Answer 2

Given that,

sec2a =3

$seca = \sqrt{3}$

secA can be written as

1+tan2A => 1+ sin2A/ cos2A

sin2A = cos2A -1

$sina \: = \sqrt{cos2a - 1}$

Which lies in first quadrant