If cosecA=2. Find 1/tanA+sinA/1+cosA
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Answered by
135
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
Anonymous:
There is another method also.
cosecA=2 1/tanA+sinA/1+cosA = 1/(sinA/cosA) +sinA/1+cosA
= (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
= (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
After cosecA = 2 in the first line, there is some space. pls don't read cosecA=2 1/tanA+sinA/1+cosA as one line.
Ur great manoharprakash.....
thanx
Answered by
29
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.....
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.....
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