JEE Main 2019 — Mathematics Trigonometry

Given ${\mathrm{cos}}^{-1}x-{\mathrm{cos}}^{-1}\frac{y}{2}=\alpha$cos−1x−cos−12y​=α

$${\mathrm{cos}}^{-1}A-{\mathrm{cos}}^{-1}B={\mathrm{cos}}^{-1}(AB+\sqrt{1-{A}^{2}}\sqrt{1-{B}^{2}}),$$cos−1A−cos−1B=cos−1(AB+1−A2​1−B2​),

$$\Rightarrow {\mathrm{cos}}^{-1}(\frac{xy}{2}+\sqrt{1-{x}^{2}}\times \sqrt{1-\frac{{y}^{2}}{4}})=\alpha$$⇒cos−1(2xy​+1−x2​×1−4y2​​)=α

$$\Rightarrow \frac{xy}{2}+\sqrt{1-{x}^{2}}\times \sqrt{1-\frac{{y}^{2}}{4}}=\mathrm{cos}\alpha$$⇒2xy​+1−x2​×1−4y2​​=cosα

$$\Rightarrow \mathrm{cos}\alpha -\frac{xy}{2}=\sqrt{1-{x}^{2}}\times \sqrt{1-\frac{{y}^{2}}{4}}$$⇒cosα−2xy​=1−x2​×1−4y2​​

Squaring both sides, we get

$${\mathrm{cos}}^{2}\alpha +\frac{{x}^{2}{y}^{2}}{4}-xy\mathrm{cos}\alpha =(1-{x}^{2})(1-\frac{{y}^{2}}{4})$$cos2α+4x2y2​−xycosα=(1−x2)(1−4y2​)

$$\Rightarrow {\mathrm{cos}}^{2}\alpha +\frac{{x}^{2}{y}^{2}}{4}-xy\mathrm{cos}\alpha =1-{x}^{2}-\frac{{y}^{2}}{4}+\frac{{x}^{2}{y}^{2}}{4}$$⇒cos2α+4x2y2​−xycosα=1−x2−4y2​+4x2y2​

$$\Rightarrow {x}^{2}-xy\mathrm{cos}\alpha +\frac{{y}^{2}}{4}=1-{\mathrm{cos}}^{2}\alpha$$⇒x2−xycosα+4y2​=1−cos2α

$$\Rightarrow {x}^{2}-xy\mathrm{cos}\alpha +\frac{{y}^{2}}{4}={\mathrm{sin}}^{2}\alpha$$⇒x2−xycosα+4y2​=sin2α

$$\Rightarrow 4{x}^{2}-4xy\mathrm{cos}\alpha +{y}^{2}=4{\mathrm{sin}}^{2}\alpha .$$⇒4x2−4xycosα+y2=4sin2α.