Math, asked by navyag689, 11 months ago

x/a cos theta + y/ b sin theta = 1 and x/a sin theta _ y/ b cos theta = 1 then prove that xsquare/ asquare + ysquare/ bsquare = 2

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Answered by dagursamiksha0004
0

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Answered by IndianInspector
1

\bigstar  According To Question ;

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

\rule{330}{2}

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

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

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

\longrightarrow\sf{(x/a\:\sin\theta - y/b\:\cos\theta)^2 = 1^2\quad...\:Equation\:4}

Add Equation 3 and Equation 4

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

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

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

When we expand Equation 3 and Equation 4 by identities, We get\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} and \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}

\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}

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

\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}

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