Q49 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
If $\cos \alpha+4 \cos \beta+9 \cos \gamma=0$ and $\sin \alpha+4 \sin \beta+9 \sin \gamma=0$, then $81 \cos (2 \gamma-2 \alpha)-16 \cos (2 \beta-2 \alpha)=$
- Correct answerA$1+8 \cos (\beta-\alpha)$
- B$\cos (\beta-\alpha)$
- C$1-36 \cos (\beta-\alpha)$
- D$1+6 \cos (\beta-\alpha)$
Solution
From given equations
$4 \cos \beta=-(\cos \alpha+9 \cos \gamma)$ ...(i)
$4 \sin \beta=-(\sin \alpha+9 \sin \gamma)$ ...(ii)
Squaring both sides and adding equations (i) and (ii)
$\begin{aligned}
& 16\left(\cos ^2 \beta+\sin ^2 \beta\right)=\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+81\left(\cos ^2 \gamma+\sin ^2 \gamma\right) \\
& +18(\cos \alpha \cdot \cos \gamma+\sin \alpha \sin \gamma) \\
& 16=1+81+18 \cos (\gamma-\alpha) \Rightarrow 9 \cos (\gamma-\alpha)=-33 \\
& \Rightarrow(-33)^2=81 \cos ^2(\gamma-\alpha)
\end{aligned}$
$\Rightarrow 1089=81 \frac{[1+\cos (2 \gamma-2 \alpha]}{2}$
$\Rightarrow 81 \cos (2 \gamma-2 \alpha)=2097$ ...(iii)
From given equations
$9 \cos \gamma=-(\cos \alpha+4 \cos \beta)$ ...(iv)
$9 \sin \gamma=-(\sin \alpha+4 \sin \beta)$ ...(v)
Squaring both sides and adding equations (iv) and (v)
$\begin{aligned}
& 81\left(\cos ^2 \gamma+\sin ^2 \gamma\right)=\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+16 \\
& \quad \quad\left(\cos ^2 \beta+\sin ^2 \beta\right)+8(\cos \alpha \cdot \cos \beta+\sin \alpha \sin \beta) \\
& \Rightarrow 81=1+16+8 \cos (\beta-\alpha) \Rightarrow \cos (\beta-\alpha)=8 \\
& \therefore \cos (2 \beta-2 \alpha)=2 \cos ^2(\beta-\alpha)-1=127 \\
& \text { Now, } 81 \cos (2 \gamma-2 \alpha)-16 \cos (2 \beta-2 \alpha) \\
& =2097-2032=65=64+1 \\
& =8 \cos (\beta-\alpha)+1
\end{aligned}$
Hence, the correct answer is option: 'A'.