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
secAcotB-secA-2cotB+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 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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