Posted by ajit yadav on Tuesday, May 14, 2013 at 1:48am.
find acute angle A and B satisfying secAcotBsecA2cotB+2=0

trigonometry  Reiny, Tuesday, May 14, 2013 at 8:38am
secAcotBsecA2cotB+2=0
(1/cosA)(cosB/sinB)  (1/cosA) 2(cosB/sinB) = 2
multiply each term by sinBcosA
cosB  sinB  2cosBcosA = 2cosAsinB
cosB  sinB  2cosBcosA + 2cosAsinB = 0
(cosB  sinB)  2cosA(cosB  sinB) = 0
(cosB  sinB)(1  2cosA) = 0
cosB = sinB OR 1  2cosA = 0
if cosB = sinB
sinB/cosB = 1
tanB = 1
B = 45° or π/4
if 12cosA = 0
cosA = 1/2
A = 60° or π/3
so A=60° , B= 45°
A=π/3 , B=π/4
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