Posted by ajit yadav on Tuesday, May 14, 2013 at 1:48am.
find acute angle A and B satisfying secAcotB-secA-2cotB+2=0
- trigonometry - Reiny, Tuesday, May 14, 2013 at 8:38am
(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 1-2cosA = 0
cosA = 1/2
A = 60° or π/3
so A=60° , B= 45°
A=π/3 , B=π/4
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