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if alpha and beta are two different values of theta lying between 0 and 2pi which satisfy the equation 6 cos theta + 8 sin theta = 9 , find the value of sin ( alpha + beta )

$Given 6 \cos θ + 8 \sin θ = 9 \Rightarrow 6 \cos θ = 9 - 8 \sin θ Squaring on both sides \Rightarrow (6 \cos θ)^{2} = {(9 - 8 \sin θ)}^{2} \Rightarrow 36 \cos^{2} θ = 81 + 64 \sin^{2} θ - 144 \sin θ \Rightarrow 36 (1 - \sin^{2} θ) = 81 - 144 \sin θ + 64 \sin^{2} θ = 0 \Rightarrow 36 - 36 \sin^{2} θ = 81 - 144 \sin θ + 64 \sin^{2} θ = 0 \Rightarrow 100 \sin^{2} θ - 144 \sin θ + 45 = 0 Since α and β are the different values of θ then Product of the roots =sin α \sin β = \frac{45}{100} = \frac{9}{20} - - - (i) 6 \cos θ + 8 \sin θ = 9 \Rightarrow 8 \sin θ = 9 - 6 \cos θ Squaring on both sides \Rightarrow (8 \sin θ)^{2} = {(9 - 6 \cos θ)}^{2} \Rightarrow 64 \sin^{2} θ = 81 + 36 \cos^{2} θ - 108 \cos θ \Rightarrow 64 (1 - \cos^{2} θ) = 81 - 108 \cos θ + 36 \cos^{2} θ = 0 \Rightarrow 64 - 64 \cos^{2} θ = 81 - 108 \cos θ + 36 \cos^{2} θ = 0 \Rightarrow 100 \cos^{2} θ - 108 \cos θ + 17 = 0 Since α and β are the different values of θ then Product of the roots =cos α \cos β = \frac{17}{100} - - - (i i) From (i) and (ii) \cos α \cos β - \sin α \sin β = \frac{17}{100} - \frac{45}{100} \Rightarrow \cos (α + β) = \frac{- 7}{25} \Rightarrow \sin (α + β) = \frac{24}{25}$

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