SOLUTION: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta b i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta

Question 1027374: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta
b
i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta
ii. Deduce that 6sin^2theta-sintheta-1=0
I have already done part i, I just found what AD, BC AND AC are in terms of their trignometric rations. I then subbed them into the equation and received it.
I don't know how to do question ii
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
6 cos +tan = 5sec

6 cos +tan = 5/cos

multiply by cos
6cos^2 +tan*cos = 5
6cos^2+sin =5
6((1-sin^2) +sin=5

6- 6sin^2 +sin=5
rearrange
6sin^2 -sin-1=0

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