Transposition Theorem in Digital Logic : Practice MCQ Questions & Answers with Explanations

$\begin{aligned} & \cos ^{-1} \mathrm{a}-\cos ^{-1} \mathrm{~b}=\cos ^{-1}\left(a b+\sqrt{1-\mathrm{a}^{2}} \cdot \sqrt{1-\mathrm{b}^{2}}\right) \\\ \therefore & \cos ^{-1} x-\cos ^{-1} \frac{y}{3} \\\ = & \cos ^{-1}\left(\frac{x y}{3}+\sqrt{1-x^{2}} \cdot \sqrt{1-\frac{y^{2}}{9}}\right)=a \\\ \therefore & \frac{x y}{3}+\frac{\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}}}{3}=\cos \alpha \\\ & x y+\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}}=3 \cos \alpha \\\ & x y-3 \cos \alpha=-\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}} \end{aligned}$ squaring on both sides, we get $x^{2} y^{2}-6 x y \cos \alpha+9 \cos ^{2} \alpha=\left(1-x^{2}\right)\left(9-y^{2}\right)$ $x^{2} y^{2}-6 x y \cos \alpha+9 \cos ^{2} \alpha=9-y^{2}-9 x^{2}+x^{2} y^{2}$ i.e., $9 x^{2}-6 x y \cos \alpha+y^{2}=9 \sin ^{2} \alpha$