If tantheta -cottheta =a and costheta +sintheta =b then 4b^2(2-b^2)

= 4(1+sin2theta)²

Step-by-step explanation:

Given the equations

tantheta -cottheta =a... 1

costheta +sintheta =b 2

We are to find 4b^2(2-b^2)

Squaring both sides of equation 2

(costheta +sintheta)^2 = b^2

cos^2theta + 2costheta sintheta + sin^2 theta = b^2

(cos^2theta+ sin^2 theta) + 2costheta sintheta = b^2

1 + 2costheta sintheta = b^2

1 + sin2theta = b^2

substitute

4b^2(2-b^2)

= 4(1+sin2theta)(2-1+sin2theta)

4(1+sin2theta)(1+sin2theta)

= 4(1+sin2theta)²

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3