Question

x/a cos theta not equals to y/b sin theta =1 and x/a sin theta-y/b cos theta=1 then prove that x square/a square+y square/b square=2​

Answer 1

Step-by-step explanation:

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Answer 2

$\bigstar$  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 $\sf{x^2/a^2 + y^2/b^2 = 2}$

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• 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

$\longrightarrow\sf{x^2/a^2\:\left(1)+y^2/b^2\:\left(1)=2}\longrightarrow\sf{x^2/a^2+y^2/b^2=2\quad...\:Hence \:Proved}$