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## Answers

cosecA=2

sinA = 1/2

A = 30 degrees

1/tanA+sinA/1+cosA = 1/tan30 +sin30/1+cos30

= √3 + (1/2) / 1+√3/2

= √3 + 1/(2+√3)

= √3 + 2-√3 [ Rationalizing 1/(2+√3) to get (2-√3) ]

1/tanA+sinA/1+cosA = 2

sinA = 1/2

A = 30 degrees

1/tanA+sinA/1+cosA = 1/tan30 +sin30/1+cos30

= √3 + (1/2) / 1+√3/2

= √3 + 1/(2+√3)

= √3 + 2-√3 [ Rationalizing 1/(2+√3) to get (2-√3) ]

1/tanA+sinA/1+cosA = 2

= (cosA/sinA) + (sinA/1+cosA)

= [cosA(1+cosA) + sinA*sinA]/ sinA(1+cosA) [ Taking LCM ]

= cosA + (cosA*cosA) + sinA*sinA]/ sinA(1+cosA)

= cosA + cos^2 A + sin^2 A/ sinA(1+cosA)

= cosA+1/ sinA(1+cosA) [ cos^2 A + sin^2 A =1]

= 1/sinA = cosecA =2

but sinA*cosecA=1

sinA=1/2....

therefore,angle A is 30

tanA=tan30=1/3^1/2

cosA=cos30=(3^1/2)/2

therefore,

1/tanA+sinA/1+cosA

=(3^1/2)+(1/2)/(1+{3^1/2}/2)

=(3^1/2)+[(1)/2+3^1/2]

by rationalising we get,

2....as the answer....

therefore,the answer is 2.....