# trigonometry

find acute angle A and B satisfying secAcotB-secA-2cotB+2=0

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1. 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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2. 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°

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3. Sin

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4. Thanks

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5. Thanks a lot. It helped me while practising for exam.

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6. Thnx Tammy om and all of u posted this sum lots of thnx

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