# maths

posted by .

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

## Similar Questions

1. ### I dont think drawing a picture will help me drwls

Which expression is equivalent to tan theta-sec theta/sin theta?
2. ### Trigonometry

I don't understand how I'm supposed set the problem up or what theta is... Use the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions. sec theta …
3. ### Precal

I got this far, but now I'm stuck...please help 6cos^2theta - 5cos(theta) + 1 = 0 6cos^2theta - 5cos(theta) = -1 cos(theta) * {6cos(theta) - 5} = -1

Identify the polar form of the linear equation 4x+3y=10. x=rcos(theta),y=rsin(theta) 4x+3y=4rcos(theta)+3rsin(theta)=10 r=10/(4cos(theta)+3sin(theta) I got it wrong
5. ### Precalculus(NEED HELP ASAP PLEASE!!)

cot(theta)= 3 pi < theta < 3pi/2 Find: sin(theta)= -1 ?
6. ### Algebra II

Multiple Choice Which expression is NOT equivalent to 1?
7. ### trig

If sin theta is equal to 5/13 and theta is an angle in quadrant II find the value of cos theta, sec theta, tan theta, csc theta, cot theta.
8. ### Calculus

Find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1. 1=5sin(theta) theta=arcsin(1/5) r'=5cos(theta) I tried the integral between arcsin(1/5) and pi-arcsin(1/5) of (((5sin(theta))^2+(5cos(theta))^2))^1/2 …
9. ### MATHS TRIGONOMETRY

if 3sin theta =2 and cos greater than 0,with the aid of a diagram determine the value of: a) tan theta b) sin theta/cos theta c) sin^2 theta + cos^2 theta
10. ### Futher maths

Tan theta +2sec theta=2 cot theta+3cosec theta=a Find theta

More Similar Questions