# maths

posted by on .

if 3 cot theta =4 find the value of 5cos theta-2sintheta/5cos theta+3sin theta

• maths - ,

3 cot ( theta ) = 4 Divide both sides by 3

cot ( theta ) = 4 / 3

sin ( theta ) = + OR - 1 / sqrt [ 1 + cot ( teta ) ^ 2 ]

In this case :

sin ( theta ) = + OR - 1 / sqrt [ 1 + ( 4 / 3 ) ^ 2 ]

sin ( theta ) = + OR - 1 / sqrt ( 1 + 16 / 9 )

sin ( theta ) = + OR - 1 / sqrt ( 9 / 9 + 16 / 9 )

sin ( theta ) = + OR - 1 / sqrt ( 25 / 9 )

sin ( theta ) = + OR - 1 / ( 5 / 3 )

sin ( theta ) = + OR - 3 / 5

cos ( theta ) = + OR - cot ( theta ) / sqrt [ 1 + cot ( teta ) ^ 2 ]

In this case :

cos ( theta ) = + OR - ( 4 / 3 ) / sqrt [ 1 + ( 4 / 3 ) ^ 2 ]

cos ( theta ) = + OR - ( 4 / 3 ) / sqrt ( 1 + 16 / 9 )

cos ( theta ) = + OR - ( 4 / 3 ) / sqrt ( 9 / 9 + 16 / 9 )

cos ( theta ) = + OR - ( 4 / 3 ) / sqrt ( 25 / 9 )

cos ( theta ) = + OR - ( 4 / 3 ) / ( 5 / 3 )

cos ( theta ) = + OR - ( 4 * 3 ) / ( 3 * 5 )

cos ( theta ) = + OR - 4 / 5

cot ( theta ) = cos ( theta ) / sin ( theta )

In quadran I cosine and sine are positive so quotient of cosine and sine will be positive.

In quadran III cosine and sine are negative so quotient of cosine and sine will be positive.

Now you have 2 sets of solutions :

1. In your equation put :

sin ( theta ) = 3 / 5

cos ( theta ) = 4 / 5

2. In your equation put :

sin ( theta ) = - 3 / 5

cos ( theta ) = - 4 / 5

5 cos ( theta ) - 2 sin ( theta ) / [ 5 cos ( theta ) ] + 3 sin ( theta )

Solutions will bee :

- 61 / 10

and

11 / 2