TS EAMCET 10 May 2024 Shift 2 Solved Paper

We start with the equation $8 \cos \theta + 15 \sin \theta = 15$. We need to find $15 \cos \theta - 8 \sin \theta$, which we will denote as $t$.First, consider squaring both sides of the given expression for clarity:Start with $(8 \cos \theta + 15 \sin \theta)^2 = 15^2$ :$64 \cos^2 \theta + 225 \sin^2 \theta + 2 \cdot 8 \cdot 15 \cos \theta \sin \theta = 225$.Simplify to:$64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta = 225$.For $t = 15 \cos \theta - 8 \sin \theta$, square both sides:$(15 \cos \theta - 8 \sin \theta)^2 = t^2$.Expanding this gives:$225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = t^2$.Now, add the two squared expressions:$64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta + 225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = 225$Simplify the combined expression:$289 (\cos^2 \theta + \sin^2 \theta) = 225 + t^2$.Since $\cos^2 \theta + \sin^2 \theta = 1$, it follows that:$289 = 225 + t^2 .$Solve for $t^2$ :$\; 289 - 225 = t^2$$\; 64 = t^2$$\; t = 8$Thus, $15 \cos \theta - 8 \sin \theta = 8$.