If x/a cos theta + y/b sin theta =1 and x/a sin theta – y/b cos theta = 1 , then prove that x2/a2 + y2/b2 =2

2 thoughts on “If x/a cos theta + y/b sin theta =1 and x/a sin theta – y/b cos theta = 1 , then prove that x2/a2 + y2/b2 =2”

1. [tex]\bigstar[/tex] According To Question ;

   • [tex]\sf{x/a\:\cos\theta + y/b\:\sin\theta = 1}[/tex] and [tex]\sf{x/a\:\sin\theta – y/b\:\cos\theta = 1}[/tex]
   • We need to Prove that [tex]\sf{x^2/a^2 + y^2/b^2 = 2}[/tex]

   [tex]\rule{330}{2}[/tex]

   • Consider [tex]\sf{x/a\:\cos\theta + y/b\:\sin\theta = 1}[/tex] as Equation 1
   • Consider [tex]\sf{x/a\:\sin\theta – y/b\:\cos\theta = 1}[/tex] as Equation 2

   Now Square Equation 1 first and Equation 2 next on Both Sides

   [tex]\longrightarrow\sf{(x/a\:\cos\theta + y/b\:\sin\theta)^2 = 1^2\quad…\:Equation\:3}[/tex]

   [tex]\longrightarrow\sf{(x/a\:\sin\theta – y/b\:\cos\theta)^2 = 1^2\quad…\:Equation\:4}[/tex]

   Add Equation 3 and Equation 4

   [tex]\longrightarrow\sf{(x/a\:\cos\theta + y/b\:\sin\theta)^2 + (x/a\:\sin\theta – y/b\:\cos\theta)^2 = 1^2+1^2}[/tex]

   Apply [tex]\sf{(a+b)^2=a^2+b^2+2ab}[/tex] identity for expanding Equation 3, Where a = [tex]\sf{x/a\:\cos\theta}[/tex] and b = [tex]\sf{y/b\:\sin\theta}[/tex] and Apply [tex]\sf{(a-b)^2=a^2+b^2-2ab}[/tex] identity for expanding Equation 4, Where a = [tex]\sf{x/a\:\sin\theta}[/tex] and b = [tex]\sf{y/b\:\cos\theta}[/tex]

   [tex]\longrightarrow\sf{(x/a\:\cos\theta + y/b\:\sin\theta)^2 + (x/a\:\sin\theta – y/b\:\cos\theta)^2 = 1+1=2}[/tex]

   When we expand Equation 3 and Equation 4 by identities, We get[tex]\sf{(x/a\:\cos\theta + y/b\:\sin\theta)^2 =\frac{x^{2}}{a^{2}} \cos ^{2} \theta+\frac{y^{2}}{b^{2}} \sin ^{2} \theta+\frac{2 x y}{a b} \sin \theta \cos \theta}[/tex] and [tex]\sf{(x/a\:\sin\theta – y/b\:\cos\theta)^2 = \frac{x^{2}}{a^{2}} \sin ^{2} \theta+\frac{y^{2}}{b^{2}} \cos ^{2} \theta-\frac{2 x y}{a b} \sin \theta \cos \theta}[/tex]

   [tex]\longrightarrow\sf{x^2/a^2\:\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+y^2/b^2\:\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=2}[/tex]

   We know [tex]\sf{\sin^2\theta+\cos^2\theta = 1}[/tex] from Trigonometric Identities

   [tex]\longrightarrow\sf{x^2/a^2\:\left(1)+y^2/b^2\:\left(1)=2}\longrightarrow\sf{x^2/a^2+y^2/b^2=2\quad…\:Hence \:Proved}[/tex]

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2. Answer:

   x/a cos theta + y/b sin theta =1 and x/a sin theta – y/b cos theta = 1

   x2/a2 + y2/b2 =2

   Step-by-step explanation:

   x2/a2 + y2/b2 =2

   Reply