Math, asked by tikschimariya111, 9 months ago

sina=root3(1-cosa) solve it​

Answers

Answered by Anonymous
1

Answer:

sina = √3 ( 1- cosa)

squaring

sin^2 a = 3 ( 1-cosa)^2

sin^2 a = 3 ( 1+ cos^2a - 2cosa)

As sin^2a = 1- cos^2 a

1- cos^2a = 3 + 3 cos^2a - 6cosa

4cos^2a - 6cos a + 2 = 0

2 cos^2 a - 3cosa +1= 0

2 cos^2a - 2cos a - cos a +1= 0

2cosa (cosa -1) -( cosa -1)=0

( 2 cos a -1)( cosa -1)=0

cosa = 1, 1/2

a = 2nΠ , 2nΠ+- Π/3

#answerwithquality #BAL

Answered by Anonymous
1

Answer:

sina = √3 ( 1- cosa)

squaring

sin^2 a = 3 ( 1-cosa)^2

sin^2 a = 3 ( 1+ cos^2a - 2cosa)

As sin^2a = 1- cos^2 a

1- cos^2a = 3 + 3 cos^2a - 6cosa

4cos^2a - 6cos a + 2 = 0

2 cos^2 a - 3cosa +1= 0

2 cos^2a - 2cos a - cos a +1= 0

2cosa (cosa -1) -( cosa -1)=0

( 2 cos a -1)( cosa -1)=0

cosa = 1, 1/2

a = 2nΠ , 2nΠ+- Π/3

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