given 7sin^2 theta + cos theta sin theta = 6 show that tan^2 theta + tan theta - 6 = 0

I worked it backwards, since the second equation was easier to solve

tan^2Ø + tanØ-6=0
(tanØ-2)(tanØ+3) = 0
tanØ = 2 or tanØ=-3

case1: tanØ=2, Ø could be in I or III
In I, sinØ = 2/√5 and cosØ = 1/√5
test in first equation
LS = 7(4/5) + (1/√5)(2/√5)
= 28/5 + 2/5 = 30/5 = 6 = RS

In III
sinØ = -2/√5 , cosØ = -1/√5
LS = 7(4/5) + (-2/√5)(-1/√5) = 30/5 = 6 = RS

Case 2: tanØ = -3, Ø could be in II or IV
in II , sinØ = 3/√10 , cosØ = -1/√10
testing in first equation,
LS = 7(9/10) + (-1/√10)((3/√10)
= 63/10 - 3/10 = 60/10 = 6 = RS

in IV, sinØ = -3/√10 , cosØ = 1/√10
LS = 7(9/10) + (1/√10)(-3/√10)
= 63/10 - 3/10 = 60/10 = 6 = RS

so all 4 values of Ø satisfy both equations.

To show that tan^2(theta) + tan(theta) - 6 = 0 using the given equation 7sin^2(theta) + cos(theta)sin(theta) = 6, we need to use some trigonometric identities and algebraic manipulation. Here's how you can do it:

1. Start with the given equation: 7sin^2(theta) + cos(theta)sin(theta) = 6.

2. Since we want to express everything in terms of tangent, we can use the Pythagorean Identity: sin^2(theta) + cos^2(theta) = 1. Rearranging this, we get sin^2(theta) = 1 - cos^2(theta).

3. Substitute the expression for sin^2(theta) into the given equation: 7(1 - cos^2(theta)) + cos(theta)sin(theta) = 6.

4. Expand the equation: 7 - 7cos^2(theta) + cos(theta)sin(theta) = 6.

5. Rearrange the terms: 7cos^2(theta) - cos(theta)sin(theta) - 1 = 0.

6. Now, let's work on expressing everything in terms of tangent. Recall that tan(theta) = sin(theta) / cos(theta).

7. Divide the entire equation by cos^2(theta): (7cos^2(theta) - cos(theta)sin(theta) - 1) / cos^2(theta) = 0 / cos^2(theta).

8. Simplify the equation using the fact that sin(theta)/cos(theta) = tan(theta): 7 - tan(theta) - 1/cos(theta) = 0.

9. Consolidate terms: 7 - tan(theta) - sec(theta) = 0.

10. Substitute the term sec(theta) with 1/cos(theta): 7 - tan(theta) - 1/cos(theta) = 0.

11. Multiply the entire equation by cos(theta) to get rid of the denominators: 7cos(theta) - sin(theta) - 1 = 0.

12. Rearrange terms: sin(theta) - 7cos(theta) + 1 = 0.

13. Now, we can use the identity tan(theta) = sin(theta) / cos(theta). Rewrite the equation in terms of tan(theta): tan(theta) - 7 + 1/cos(theta) = 0.

14. Substitute the expression 1/cos(theta) with sec(theta): tan(theta) - 7 + sec(theta) = 0.

15. Finally, rearrange the terms to match the desired equation: tan^2(theta) + tan(theta) - 6 = 0.

Therefore, we have successfully shown that tan^2(theta) + tan(theta) - 6 = 0 using the given equation 7sin^2(theta) + cos(theta)sin(theta) = 6.